Post before! Anyway, my solution

$x=1, y=1: f(1)f(f(1))+f(f1))=f(1)f(f(1))+f(1)f(f(1))\implies \\ \implies f(1)f(f(1))=f(f(1))$, but as $f(f(1))\in \mathbb{R}^{+}\implies f(1)=1$ $x=1, y=x: f(f(x))+f(xf(1))=f(x)+f(x)f(f(x^2))\implies$ $$f(f(x))=f(x)f(f(x^2))$$$y=1: xf(x^2)=f(x)\Longleftrightarrow \boxed{x^2f(x^2)=xf(x)}$ $x,x: xf(x^2)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2))=2\frac{f(x)}{x}f(f(x^2))=2\frac{f(f(x))}{x}$ or $xf(x)f(f(x))+xf(xf(x))=2f(f(x))$ Put $x\rightarrow x^2$ in the above and get $x^2f(x^2)f(f(x^2))+x^2f(x^2f(x^2))=2f(f(x^2))\Longleftrightarrow$ $\Longleftrightarrow xf(xf(x))=f(f(x))(2-xf(x))\\ \text{and} \\ x^2f(x^2f(x^2))=f(f(x^2))(2-xf(x))\implies \\ \frac{x}{f(f(x))}=\frac{x^2}{f(f(x^2))}\implies f(x)=\frac{1}{x}$ for any positive real number $x$

As usual, let $P(x,y)$ denote the problem condition when $(x,y)$ is plugged in. $P(1,1)$ is $f(1)f(f(1))+f(f(1))=2f(1)f(f(1))$, so $f(1)=1$. $P(1,x)$ and $P(x,1)$ are \begin{align*}f(f(x))+f(x) &= f(x)\left(1+f(f(x^2))\right),\\ xf(x^2)+f(f(x)) &= f(x)\left(f(f(x^2))+1\right), \end{align*}which reduce to \begin{align} f(f(x))&=f(x)f(f(x^2)),\\ xf(x^2)&=f(x), \end{align}where we used $(1)$ to get $(2)$. $P(x,1/x)$ is $$xf(x^2)f\left(f\left(\frac{1}{x}\right)\right) + f\left(\frac{f(x)}{x}\right) = f(f(x^2)) + f\left(f\left(\frac{1}{x^2}\right)\right),$$so $$f(x)f\left(f\left(\frac{1}{x}\right)\right) = f\left(f\left(\frac{1}{x^2}\right)\right) = \frac{f\left(f\left(\frac{1}{x}\right)\right)}{f\left(\frac{1}{x}\right)}.$$Therefore, $f(x)f(1/x)=1$ for all $x$. Using $(2)$, rewrite the problem condition as $$f(x)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2))+f(f(y^2))\right).$$$P(x^2,x)$ gives \begin{align*} f(x^2)f(f(x)) + f(f(x)) &= f(x^3) \left(f(f(x^4))+f(f(x^2))\right)\\ &= f(x^3) \left(\frac{f(f(x^2))}{f(x^2)}+\frac{f(f(x))}{f(x)}\right)\\ &= f(f(x))f(x^3) \left(\frac{1}{f(x^2)f(x)}+\frac{1}{f(x)}\right)\\ &= \frac{f(f(x))f(x^3)(f(x^2)+1)}{f(x^2)f(x)}. \end{align*}Therefore, $f(x^2)f(x)=f(x^3)$. $P(x,x)$ is $$f(x)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2)).$$Note that \begin{align*} 0 < f(xf(x)) &= 2f(x^2)f(f(x^2)) - f(x)f(f(x))\\ &= 2f(x^2)f(f(x^2)) - f(x)^2f(f(x^2))\\ &= 2f(x^2)f(f(x^2)) - xf(x^2)f(x)f(f(x^2))\\ &= f(x^2)f(f(x^2))(2-xf(x)), \end{align*}so $xf(x)<2$ for all $x$. Let $g(x)=xf(x)$, and note that $f(x)f(1/x)=1$ implies $g(x)g(1/x)=1$. I claim that $g(x^3)=g(x)^2$. Note that $f(x)=\frac{g(x)}{x}$, and by $(2)$, $f(x^2)=\frac{g(x)}{x^2}$. Therefore, $f(x^3)=f(x)f(x^2)=\frac{g(x)^2}{x^3}$, proving the claim. Suppose $g(x)>1$ for some $x$. Then $g(x^{3^n})=g(x)^{2^n}>2$ for sufficiently large integers $n$, a contradiction. If $g(x)<1$, then $g(1/x)>1$, which gives a contradiction as before. Therefore, $g(x)=1$ for all $x$, so $\boxed{f(x)=\frac{1}{x}}$ is the only possible solution. We can easily check that this works.

My solution If $f$ is constant then $f \equiv 0$ which is a contradiction If $f$ is nonconstant , let $P(x_0,y_0)$ be the assertion $P(1,1) : f(f(1)) = f(1).f(f(1)) \implies f(1) = 1$ $P(x,1) : xf(x^2)+f(f(x)) = f(x) ( f(f(x^2))+1)$ $P(1,y) : f(f(y))+f(y) = f(y)(f(f(y^2))+1) \implies f(x)+f(f(x)) = f(f(x))+xf(x^2) \implies f(x^2) = \frac{f(x)}{x}$ Once again , $P(x,1)$ gives $f(x)+f(f(x)) = f(x)+f(x)f(f(x^2)) \implies f(f(x^2)) = \frac{f(f(x))}{f(x)} = f(\frac{f(x)}{x}) = f(f^2(x))$ Now we prove $f$ is injective . Suppose that there exist $f(a) = f(b) = c$ From the conditions we have $f(x)f(f(y)) + f(yf(x)) =xf(x^2)f(f(y)+f(yf(x)) = f(xy)(f(f(x^2))+f(f(y^2))) = f(xy).(f(f^2(x)+f(f^2(y)))$ Now we set $x = a , x = b$ respectively ; then , we receive $f(ay) = f(by) \forall $ $y \in \mathbb R^{+}$ Then $f(a^2) = f(ab) = f(b^2) = \frac{f(a)}{a} = \frac{f(b)}{b} = \frac{c}{a} = \frac{c}{b} \implies a = b$, which is a contradiction Then $f$ is injective . From above we almost have $f(f(x^2)) = f(f^2(x)) \implies f^2(x) = f(x^2) = \frac{f(x)}{x} \implies f(x) = \frac{1}{x}$ Then , the solution is $f(x) = \frac{1}{x} $ $\forall$ $x $ $\in $ $\mathbb R^{+}$

mathwizard888 wrote: Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$, $$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$$ Answer: The only functions which work are $f(x)=\tfrac{1}{x}$ for all $x>0$. Let $P(x, y)$ denote the given functional equation: $$xf(x^2)f(f(y))+f(yf(x))=f(xy)\cdot \left(f(f(x^2))+f(f(y^2))\right),$$then it is easy to see that $f(x) \equiv \tfrac{1}{x}$ is indeed a solution. We now show that this is the only one. Note that $$P(1, 1) \implies f(1)=1$$and $$P(1, y) \implies f(f(y))=f(y)f(f(y^2)),$$for all $y>0$. Also, $$P(x, 1) \implies xf(x^2)=f(x),$$for all $x>0$. Hence, we can rewrite $P$ as $$f(x)f(f(y))+f(yf(x))=f(xy) \cdot \left(\frac{f(f(x))}{f(x)}+f(f(y^2))\right).$$ Claim: $f$ is injective. (Proof) Suppose $f(a)=f(b)$ for some $a \ne b$. Together, $P(a, y)$ and $P(b, y)$ yield that $f(ay)=f(by)$ for all $y>0$; hence, for $t=\tfrac{1}{a}$ we have $f(t)=f(1)=1$. However, this yields $f(t^2)=f(t \cdot t)=f(t)=1,$ so $$f(t)=tf(t^2)=tf(t) \implies t=1,$$hence $\boxed{a=b}$. $\blacksquare$ Finally, we observe that $$f\left(f(x)^2\right)=\frac{f(f(x))}{f(x)}=f(f(x^2))=f\left(\frac{f(x)}{x}\right) \implies f(x)^2=\frac{f(x)}{x} \implies f(x) \equiv \frac{1}{x},$$due to injectivity, hence we conclude that the initial assertion is valid. $\blacksquare$

The only such function is $f(x) = \frac{1}{x}$ for all $x \in \mathbb{R}^{+}$, which is easily seen to satisfy the given functional equation. Let $P(x, y)$ denote the usual. Using $P(1, 1)$, we get $f(1)f(f(1)) + f(f(1)) = 2f(1)f(f(1))$, and hence $f(1) = 1$. Now, using $P(1, x)$, we get $$f(f(x)) + f(x) = f(x)f(f(x^2)) + f(x)$$$$\implies$$ $$\boxed{f(f(x)) = f(x)f(f(x^2))} \quad (1)$$ Thus, using $P(x, 1)$ we obtain: $$xf(x^2) + f(f(x)) = f(x)f(f(x^2)) + f(x) = f(x) + f(f(x))$$$$\implies$$ $$\boxed{xf(x^2) = f(x)} \quad (2)$$ We now show that $f(t) = 1$ implies $t = 1$. In this case then using $(2)$ and $(1)$ with $x = t$ we get $f(t^2) = \frac{1}{t}$ and $f\left(\frac{1}{t}\right) = 1$. Then $P(t, t)$ gives $$t \cdot \frac{1}{t} + 1 = \frac{2}{t}$$$$\implies$$$$t = 1$$ As desired. We now proceed to show that $f$ is injective. Assume $f(a) = f(b)$, then $(1)$ and $(2)$ give $af(a^2) = bf(b^2)$ and $f(f(a^2)) = f(f(b^2))$, so comparing $P(a, x)$ and $P(b, x)$ gives $f(ax) = f(bx)$ for all positive real $x$. Take $x = \frac{1}{a}$ to get $f\left(\frac{b}{a}\right) = f(1) = 1$, and hence $a = b$ by the previously shown statement. Thus $f$ is injective. Finally, plug in $f(x)$ in $(2)$ and compare with $(1)$ to obtain $$f(x)f(f(x)^2) = f(f(x)) = f(x)f(f(x^2))$$$$\implies$$$$f(f(x)^2) = f(f(x^2))$$$$\implies$$$$f(x)^2 = f(x^2)$$ Finally, plug this relation into $(2)$ to obtain $f(x) = \frac{1}{x}$ for all positive real $x$.

My solution: Let $P(x,y)$ denote the given assertion. $$P(1,1) \Rightarrow f(1)f(f(1))+f(f(1))=2f(1)f(f(1) \Rightarrow f(1)=1$$$$P(1,x) \Rightarrow f(f(x))+f(x)=f(x)(1+f(f(x^2))) \Rightarrow f(f(x))=f(x)f(f(x^2)) \text{ } \text{ } \text{ } -(A)$$$$P(x,1) \Rightarrow xf(x^2)+f(f(x))=f(x)f(f(x^2))+f(x)=f(f(x))+f(x) \Rightarrow f(x)=xf(x^2) \text{ } \text{ } \text{ } -(B) \text{ } \text{ } \text{ } \text{[Using }A \text{]}$$On putting $x \rightarrow f(x)$ in $B$, we get that $f(f(x))=f(x)f(f(x)^2)$. Then, using $A$, we have $f(f(x)^2)=f(f(x^2))$ $-(C)$. $$P(x,x) \Rightarrow xf(x^2)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2)) \Rightarrow f(x)f(f(x))+f(xf(x))=2f(x^2)f(f(x^2)) \text{ } \text{ } \text{ } -(D) \text{ } \text{ } \text{ } \text{[Using }B \text{]}$$ Now, Let $f(a)=f(b)$ for some $a,b \in \mathbb{R^+}$. Then, from $C$, we have $f(f(a^2))=f(f(b^2))$. Also, From $P(a,y)$ and $P(b,y)$, along with $B$, we get that $f(ay)=f(by) \text{ } \forall y \in \mathbb{R^+}$. Putting $y=f(a)=f(b)$ above, we have $f(af(a))=f(bf(a))=f(bf(b))$. Thus, On putting $x=a$ and $x=b$ in $D$, we get that $f(a^2)=f(b^2)$. But then, from $B$, we have $a=b$. This means that $f$ is injective. Finally, Using injectivity on $C$, we have $f(x^2)=f(x)^2$. And with the help of $B$, we get $f(x)=xf(x^2)=xf(x)^2 \Rightarrow f(x)=\frac{1}{x} \text{ } \forall x \in \mathbb{R^+}$

Here's a solution that doesn't use injectivity. Let $P(x,y)$ be the given assertion. $P(1,1) : f(1)f(f(1)) + f(f(1)) = 2f(1)f(f(1))$, so we have $f(1)=1$ Combining $P(1,x)$ and $P(x,1)$ gives $xf(x^2)=f(x)...(1)$. We rewrite the original equation as $$f(x)f(f(y))+f(yf(x))=f(xy)\left(\frac{f(f(x))}{f(x)}+\frac{f(f(y))}{f(y)}\right)$$$P(x,\frac{1}{x}) : f(x)f(f(\frac{1}{x}))+f(\frac{f(x)}{x})=f(f(x^2))+\frac{f(f(\frac{1}{x}))}{f(\frac{1}{x})}=f(\frac{f(x)}{x})+\frac{f(f(\frac{1}{x}))}{f(\frac{1}{x})}\Longrightarrow f(x)f(\frac{1}{x})=1$ $P(x,x) : f(x)f(f(x))+f(xf(x))=\frac{2f(x^2)}{f(x)}f(f(x))=\frac{2}{x}f(f(x))...(2)$ $P(x,\frac{1}{xf(x)}) : f(x)f(f(\frac{1}{xf(x)}))+f(\frac{1}{x})=f(\frac{1}{f(x)})\left(\frac{f(f(x))}{f(x)}+\frac{f(f(\frac{1}{xf(x)}))}{f(\frac{1}{xf(x)})}\right)=f(\frac{1}{x})+f(\frac{1}{f(x)})\frac{f(f(\frac{1}{xf(x)}))}{f(\frac{1}{xf(x)})}$ $\Longrightarrow f(x)f(f(x))=f(xf(x))=(\frac{2}{x}-f(x))f(f(x)) (\because (2))\Longrightarrow f(x)=\frac{1}{x}$. Hence $\boxed{f(x)=\frac{1}{x}}$ is the only solution.

Run of the mill FE. Not easy, but standard. The only solution is $f(x)=1/x$. It is easy to see that this works. Now, suppose $f$ is the function satisfying the FE, and let $P(x,y)$ be the statement of the FE. We see that \[P(1,1)\implies f(1)f(f(1))+f(f(1))=f(1)\cdot 2f(f(1))\implies f(1)+1=2f(1)\implies f(1)=1.\]Note that we can cancel stuff since the domain and codomain exclude $0$. We now see that \begin{align*} P(x,1) &\implies xf(x^2)+f(f(x)) = f(x)f(f(x^2))+f(x) \\ P(1,x) &\implies f(f(x))+f(x)=f(x)f(f(x^2))+f(x). \end{align*}Thus, $f(x^2)=f(x)/x$, and $f(f(x^2))=f(f(x))/f(x)$. Using this, we simplify the FE to \[f(x)f(f(y))+f(yf(x))=f(xy)\left(\frac{f(f(x))}{f(x)}+\frac{f(f(y))}{f(y)}\right),\]and call the above statement $Q(x,y)$. Note that \[Q(x,x)\implies f(x)f(f(x))+f(xf(x))=f(x^2)\cdot\frac{2f(f(x))}{f(x)}=\frac{2f(f(x))}{x}.\]Thus, \[f(xf(x))=f(f(x))\left(\frac{2}{x}-f(x)\right).\] Claim: $f$ is injective. Proof of Claim: Suppose $f(a)=f(b)=k$. Then, $Q(a,b)$ implies \[kf(k)+f(bk)=f(ab)\cdot 2\frac{f(k)}{k}.\]But \[f(bk)=f(bf(b))=f(k)(2/b-k),\]so \[\frac{2f(k)}{b}=\frac{2f(k)}{k}f(ab).\]Thus, $b=k/f(ab)$. Similarly, we get $a=k/f(ab)$, so $a=b$. Thus, $f$ is injective. $\blacksquare$ Now, we have \[f(f(x^2))=f(f(x))/f(x)=f(f(x)^2),\]so $f(x^2)=f(x)^2$. Thus, $f(x)^2=f(x)/x$, which means that $f(x)=1/x$, as desired. $\blacksquare$

Denote the assertion as $P(x,y)$. $P(1,1)$ yields $f(1)f(f(1))+f(f(1))=2f(1)f(f(1))\implies f(1)+1=2f(1)$, so $f(1)=1$. $P(1,x)$ gives $f(f(x))=f(f(x^2))f(x)$ and $P(x,1)$ yields $xf(x^2)=f(x)\,\,\,(*)$. Now, consider the two equations: $P(x,y)$ and $P(y,x)$. As the RHS is symmetric, these two have the same LHS. In particular, $$f(x)f(f(y))+f(yf(x))=f(y)f(f(x))+f(xf(y))$$Suppose that $f(a)=f(b)$. Substituting $(a,b)$ into the equation above, we get that $f(af(b))=f(bf(a))$. Now, considering $P(a,b)$ and $P(a,a)$, it is not hard to see that the LHS is the same for both, as well as the second factor of the RHS, so we have $f(ab)=f(a^2)$. Similarly, $f(ab)=f(b^2)$, so $f(a^2)=f(b^2)$. Now, compare $f(a,b)$ and $f(b,a)$. The RHS is the same for both, as well as the latter term on the LHS, so we have that $$af(a^2)f(f(b))=bf(b^2)f(f(a))$$and therefore $a=b$. So, $f$ is injective. To finish, note that if we plug in $f(x)$ into $(*)$, we get $f(x)f(f(x)^2)=f(f(x))$, so $f(f(x)^2)=f(f(x^2))$, and injuectivity gives $f(x^2)=f(x)^2$. Using this in $(*)$ will give $xf(x)=1$, or $f(x)=\frac{1}{x}$, so this is the only solution.

We claim that $f(x) \equiv 1/x$ is the only solution. By substitution, we see that it works. Let $P(x, y)$ be the assertion in the problem statement. $P(1, 1)$ gives $f(1) = 1$. $P(x, y)$ combined with $P(y, x)$ gives: $f(x^2) = f(x)/x$ - $(1)$ $f(f(x)) = f(x)f(f(x^2))$ - $(2)$ Now, suppose $f(a) = f(b)$. Then, $P(a, y)$ combined with $P(b, y)$ combined with $(1)$ and $(2)$ gives $f(ay) = f(by)$ which gives $f(x) = f(b/a * x)$. Then, $(1)$ gives $f(b/a) = f(b/a*b/a)*b/a = f(b/a)*b/a$ which gives $a = b$. Thus, $f$ is injective. Considering $(2)$, $f(f(x^2)) = f(f(x))/f(x) = f(f(x)^2)$ by $(1)$, and applying injectivity gives $f(x^2) = f(x)^2$. Putting this in $(1)$ gives $f(x) = 1/x$, as desired.

We claim that $f(x) = \frac{1}{x}$ is the only solution. It is easy to check that this function indeed works. Now, we will prove that it is the only one. Let $P(x,y)$ denote the assertion of $x$ and $y$ to the given functional equation. Now, notice that $P(1,1)$ gives us $f(1) = 1$. Furthermore, $P(x,1)$ and $P(1,x)$ gives us: \[ xf(x^2) + f(f(x)) = f(x) (f(f(x^2)) + 1) \]\[ f(f(x)) + f(x) = f(x) ( f(f(x^2)) + 1) \]From here, we get that $f(f(x)) = f(x) f(f(x^2)) $ and $f(x) = xf(x^2)$. Now, notice that the given functional equation is now equivalent to \[ f(x)f(f(y)) + f(yf(x)) = f(xy) \left( \frac{f(f(x))}{f(x)} + \frac{f(f(y))}{f(y)} \right) \]Take $P(x,x)$ now : \[ f(x)f(f(x)) + f(xf(x)) = f(x^2) \cdot \frac{2f(f(x))}{f(x)} = \frac{2f(f(x))}{x} \]Change $x$ to $x^2$ in the equation above, and by simplifying: \[ \frac{f(f(x))}{x} + f(xf(x)) = \frac{2f(f(x)))}{x^2 f(x)} \]Comparing last two equations we have, \[ f(f(x)) \left( \frac{2}{x} - f(x) \right) = f(f(x)) \left( \frac{2}{x^2f(x)} - \frac{1}{x} \right) \]\[ \frac{3}{x} = f(x) + \frac{2}{x^2 f(x)} \]Solve this give us $f(x) = \frac{1}{x}$ or $f(x) = \frac{2}{x}$. Checking back to the equation, the one satisfy is only $f(x) = \frac{1}{x}$.

This proof is basically the proof above, but i'm an arrogant kid that wants to flex his math skills so this i'm leaving this for storage. First of all we claim that $f(x) = \dfrac{1}{x}$ is the only solution. This clearly works. Also, denote the assertion as $P(x,y)$ $P(1, 1)$ gives us that $f(1) = 1$. Plugging $P(1,x)$ then gives us our first major observation: \begin{align*} f(f(x^2)) &= \dfrac{f(f(x))}{f(x)} \end{align*}In addition, plugging in $P(x,1)$ gives us that: \begin{align*} xf(x^2) + f(f(x)) &= f(x) + f(f(x)) \\ f(x^2) &= \dfrac{f(x)}{x} \end{align*}Note that this simplifies the function equation into: \begin{align*} f(x)f(f(y)) + f(yf(x)) &= f(xy) \sum_{cyc} \dfrac{f(f(x))}{f(x)} \end{align*}Also note that we have: \begin{align*} f( \frac{f(x)}{x} ) &= f(f(x^2)) \\ &= \dfrac{f(f(x))}{f(x)} \end{align*}This finally gives us that plugging in $P(x, \dfrac{1}{x})$ and simplifying, we have: \begin{align*} f(\frac{1}{x}) &= \dfrac{1}{f(x)} \end{align*}Our final step now is to plug in $P(x,x)$. We have that after some algebra we find: \begin{align*} f(xf(x)) &= f(f(x)) \left (\dfrac{2}{x} - f(x) \right ) \end{align*}Now, since $f(x^2) = \dfrac{f(x)}{x}$, we have that $xf(x) = x^2f(x^2)$. This gives us that by plugging in $x^2$ into the equation above and equating, we find \begin{align*} f(x) &= \dfrac{1}{x} \thickspace \text{or} \thickspace \dfrac{2}{x} \end{align*}for all real $x$. Finally using the fact that $f(\frac{1}{x}) = \dfrac{1}{f(x)}$ gives us that $\boxed{f(x) = \dfrac{1}{x}}$ for all positive reals $x$.

My Solution Let $P (x,y) $ be the assertion of $xf (x^2)f (f (y))+f (y f (x))=f (xy)(f (f (x^2))+f (f (y^2)) $ $P (1,1) $ gives $f (1)=1$ hence, $P (1,x) $ shows $f (f (x))=f (x)f (f (x^2))\cdots (1) $ From $P (x,1) $ we get $xf (x^2)=f (x) $ or $f (x^2)=\frac {f (x)}{x}\cdots (2) $ Now, $P\left (x,\frac {1}{x}\right)=f (x)f (f \left(\frac {1}{x}\right)=f (f \left(\frac {1}{x^2})\right) $ Finally, $f (x)f (\frac {1}{x})=1$ Applying $P \left(\frac {1}{x},\frac {1}{f (x)}\right)=1$ $\frac {1}{xf (x^2)f (f (f (x)))}+\frac {1}{f (f (x)^2)}=\frac {1}{f (xf (x))}\left (\frac {1}{f (f (x^2))}+\frac {1}{f (f (f (x^2)))}\right) $ Simplifying that, $f (xf (x))^2 f (f (x))=f (x)^2 f (f (x))f (f (x)^2) $ Simplifying $(1) $ and $(2) $ $f (x f (x))^2=f (x)f (f (x)) $ $P\left (\frac {1}{x},\frac {1}{x}\right)\implies f (f (x))=x^4f (x)^3$ Note that, $x^4f (x)^3=f (f (x))=f (x)f (f (x^2))=f (x)x^8f (x^2)^3=x^5f (x)^4$ Therefore, $\boxed {f (x)=\frac {1}{x}}$ $ \forall x\in\mathbb {R^+} $

Posting for storage; solved with TheUltimate123. The answer is $f(x) = \tfrac{1}{x}$, which can be verified to work. Let $P(x,y)$ denote the assertion. We prove a few preliminary facts. By $P(1,1)$, we receive $$f(1)f(f(1)) + f(f(1)) = f(1)(f(f(1)) + f(f(1))) \implies f(1) + 1 = 2 \implies f(1) = 1.$$ By $P(x,1)$ and $P(1,x)$, we get $$xf(x^2) + f(f(x)) = f(x)(f(f(x^2)) + 1) = f(f(x)) + f(x) \implies f(x) = xf(x^2).$$ By $P(\tfrac{1}{x},x)$ we get $$\frac{1}{x}f\left(\frac{1}{x^2}\right)f(f(x)) + f\left(xf\left(\frac{1}{x}\right)\right) = f(f(x^2)) + f\left(f\left(\frac{1}{x^2}\right)\right) \implies f\left(\frac{1}{x}\right) f(f(x)) = f(f(x^2)).$$Substituting this into $P(1,x)$, we get $$f(f(x)) + f(x) = f(x)(f(f(x^2)) + 1) = f(x)f\left(\frac{1}{x}\right)f(f(x)) + f(x) \implies f(x)f\left(\frac{1}{x}\right) = 1.$$ We now show that $f$ is injective. Say $f(a) = f(b)$. Then $P(a,b)$ and $P(b,a)$ show $$af(a^2)f(f(b)) + f(bf(a)) = f(ab)(f(f(a^2))+f(f(b^2))) = bf(b^2)f(f(a)) + f(af(b)),$$which actually implies $f(a)f(f(b)) + f(bf(a)) = f(b)f(f(a)) + f(af(b))$, or $f(af(b)) = f(bf(a))$. Moreover, $$f(f(a^2)) = f(f(a))f\left(\frac{1}{a}\right) = \frac{f(f(a))}{f(a)} = \frac{f(f(b))}{f(b)} = f(f(b))f\left(\frac{1}{b}\right) = f(f(b^2)).$$Then consider $P(a,a)$ and $P(b,b)$. Then the LHSes and the second factors of the RHSes are equal, so $f(a^2) = f(b^2)$. But $af(a^2) = f(a) = f(b) = bf(b^2)$, so $a=b$. Finally, note $$f(f(x)^2) = \frac{f(f(x))}{f(x)} = f\left(\frac{1}{x}\right)f(f(x)) = f(f(x^2)),$$id est $f(x)^2 = f(x^2) = f(x)/x \implies f(x) = \tfrac{1}{x}$. The end.

mathwizard888 wrote: Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$, $$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$$ The only function that works is $\boxed{\tfrac{1}{x}}$. Now we'll prove this is the only. Claim 1 $f(1)=1$ Proof$\; \text{P}(1,1)\implies f(f(1))(f(1)-1)=0\implies f(1)=1 \;$ since $f(f(1))\ne 0$. Claim 2$f(f(y^2))=\dfrac{f(f(y))}{f(y)}$ Proof $\; \text{P}(1,y)\implies f(f(y))+f(y)=f(y)(1+f(f(y^2))) \implies f(f(y^2))=\dfrac{f(f(y)}{f(y)}$. Claim 3 $f(x^2)=\dfrac{f(x)}{x}$ Proof$\;\text{P}(x,1)\implies xf(x^2)+f(f(x))=f(x)(1+f(f(x^2)))=f(x)+f(f(x))$ by Claim 2.Hence $xf(x^2)=f(x)\implies f(x^2)=\dfrac{f(x)}{x}$. Claim 4 $f(a)=f(b)\implies f(ay)=f(by)$ Proof$\;$Comparing $\text{P}(a,y),\text{P}(b,y)$ concludes the Claim. Claim 5$f$ is injective Proof$\;$Let $a,b\in \mathbb{R^+} \; a\ne b$.Let $\varepsilon=\tfrac{a}{b}$.So by Claim 4 we get $f(y)=f(\tfrac{a}{b}\cdot y)=f(\varepsilon y)$ Thus $$f(y^2)=f(\varepsilon y^2)=f(\varepsilon^2 y^2)=\dfrac{f(\varepsilon y)}{\varepsilon y}=\dfrac{f(y)}{\varepsilon y}=\dfrac{f(y^2)}{\varepsilon}\implies \varepsilon =1$$. where we have used Claim 2 and Claim 3 repeatedly.Hence $f$ is injective. Now $$f(f(x)^2)=\dfrac{f(f(x))}{f(x)}=f(f(x^2))\implies f(x)^2=f(x^2)=\dfrac{f(x)}{x}\implies f(x)=\dfrac{1}{x}$$.$\blacksquare$

Let $P(x,y)$ denote the assertion. $P(1,1)$ gives $f(f(1))=f(1)f(f(1)).$ Since $f$ cannot be zero, $f(1)=1$. Thus, $P(x,1)$ and $P(1,y)$ become \begin{align*} xf(x^2)+f(f(x)) &= f(x)f(f(x^2))+f(x) \\ f(f(y)) &= f(y)f(f(y^2)) \end{align*}respectively. By the second one, we have $xf(x^2)=f(x)$. We have \begin{align*} \frac{f(f(x))}{f(x)} &= f(f(x^2)) \\ \frac{f(x)}{x} &= f(x^2). \end{align*}Thus, $f(f(x)^2)=f(f(x^2))$. We claim that $f$ is injective. Suppose $f(a)=f(b)$. Taking $P(a,y)$ we get \[af(a^2)f(f(y)) + f(yf(a)) = f(ay) \left(f(f(a^2)) + f(f(y^2))\right)\]Now, $f(a)f(f(a))=f(b)f(f(b))\implies f(f(a^2))=f(f(b^2))$, so while $a$ changes between $a$ and $b$, $f(f(a^2))+f(f(y^2))$ is constant. $f(yf(a))$ is clearly constant, and so is $af(a^2)=f(a)$. Therefore, $f(ay)$ is constant. Thus, $f(a^2)=f(ab)=f(b^2)$, so $f$ is injective as desired. Therefore, $f(x)^2=f(x^2).$ Now, use $xf(x^2)=f(x)$ to get $f(x)=1/x$ which works.

probably overcomplicated things a little bit

Let $P(x,y):=xf(x^2)f(f(y)) + f(yf(x)) = f(xy)(f(f(x^2)) + f(f(y^2))$ $P(1,1)$ yields $f(1)f(f(1))+f(f(1))=f(1)(2f(f(1)))\Longrightarrow f(f(1))(f(1)+1)=2f(1)f(f(1))\overset{\text{since the function is positive}}{\Longrightarrow} f(1)+1=2f(1)\implies f(1)=1$ $P(x,1)$ yields $xf(x^2)+f(f(x))=f(x)(f(f(x^2))+1)$ (1) $P(1,x)$ yields $f(f(x))+f(x)=f(x)(f(f(x^2))+1)$, which can also be simplified into $f(f(x))=f(x)f(f(x^2))$ (2) Thus, from (1) and (2), $xf(x^2)+f(f(x))=f(x)+f(f(x))\Longrightarrow f(x)=xf(x^2)$ (!) Furthermore, substituting $x\to f(x)$ in (!) yields $f(f(x))=f(x)f(f(x)^2)$, thus from (2), $f(x)f(f(x)^2)=f(x)f(f(x^2))\Longrightarrow f(f(x)^2)=f(f(x^2))$ Claim: The function is injective Proof: Let $f(a)=f(b)$ $P(a,y)$ yields $af(a^2)f(f(y))+f(yf(a))=f(ay)(f(f(a^2))+f(f(y^2)))=f(ay)(f(f(a)^2)+f(f(y^2)))$ $P(b,y)$ yields $bf(b^2)f(f(y))+f(yf(b))=f(by)f(f(b^2))+f(f(y^2)))=f(ay)(f(f(a)^2)+f(f(y^2)))$ furthermore comparing the two, we get that $f(ay)=f(by), \forall a,b,y \in (0,\infty)$ now let $f(ay)=f(by):=Q(a,b,y)$ $Q(a,b,\frac{1}{a})$ yields $f\left(\frac{b}{a}\right)=f(1)=1$ which implies that $a=b$, thus the function is injective $\square$. Thus $f(f(x)^2)=f(f(x^2))\overset{\text{from injectivity}}{\Longrightarrow} f(x)^2=f(x^2)$, now we can rewrite (!) as: $f(x)=xf(x^2)=xf(x)^2\Longrightarrow f(x)=xf(x)^2\Longrightarrow xf(x)=1$ $\therefore f(x)=\frac{1}{x}$ $\blacksquare$. So to sum up, $\boxed{f(x)=\frac{1}{x}, \forall x \in (0,\infty)}$

mathwizard888 wrote: Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$, $$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$$ $P(1,1)$ gives $f(1)=1$ $P(1,y)$ gives $f(f(y))=f(y)f(f(y^2))$ (1) $P(x,1)$ gives $xf(x^2)+f(f(x))=f(x)f(f(x^2))+f(x)$ (2) (1),(2) gives $f(x)=xf(x^2)$ (3) In (3) for $x\rightarrow f(x)$ we have using (1) : $f(f(x)^2)=f(f(x^2))$ (4) Now we are going to prove that $f:1-1$. Using (3) the first can be written as: $f(x)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$ Let $f(a)=f(b)$ then: $P(a,b),P(b,a)$ gives $f(af(b))=f(bf(a))$ Now $P(a,a)P(b,b)$ using the last one gives: $f(a^2)f(f(a^2))=f(b^2)f(f(b^2))\Rightarrow f(a^2)f(f(a)^2)=f(b^2)f(f(b)^2)\Rightarrow f(a^2)=f(b^2)\Rightarrow \frac{f(a)}{a}=\frac{f(b)}{b}\Rightarrow a=b$ we use (4) and (1). Now because $f$ is ingective (4) can be writen as: $f(x)^2=f(x^2)$ (5) and using (5) in (1) we have $f(y)f(f(y))=1$ (6) $P(x,x)$ with (6) gives: $1+f(xf(x))=2\Rightarrow f(xf(x))=1=f(1)\Rightarrow xf(x)=1\Rightarrow f(x)=\frac{1}{x}$

The answer is $f(x)=\tfrac{1}{x}$ only, which clearly works. Let $P(x,y)$ denote the assertion. From $P(1,1)$, we find that $f(f(1))=f(1)f(f(1))$, hence $f(1)=1$. From $P(1,x)$ we find that $$f(f(x))+f(x)=f(x)(1+f(f(x^2))) \implies f(f(x))=f(x)f(f(x^2)),\qquad(1)$$and from $P(x,1)$ we find that $$xf(x^2)+f(f(x))=f(x)(f(f(x^2))+1) \implies xf(x^2)=f(x),\qquad(2)$$using $(1)$. I now claim that $f$ is injective. Suppose that $f(a)=f(b)$ where $a \neq b$. By comparing $P(a,1)$ and $P(b,1)$ and using the fact that $f(f(a^2))=\tfrac{f(f(a))}{f(a)}=\tfrac{f(f(b))}{f(b)}=f(f(b^2))$, we find that $af(a^2)=bf(b^2)$. Using this, by comparing $P(a,x)$ and $P(b,x)$ it follows that $f(ax)=f(bx)$ for all $x$. Hence $f(a^2)=f(ab)=f(b^2)$ by applying this twice, which contradicts $af(a^2)=bf(b^2)$, hence proven. Finally, from substituting $f(x)$ into $(2)$ and using $(1)$, it follows that $$f(f(x)^2)=\frac{f(f(x))}{f(x)}=f(f(x^2)) \implies f(x)^2=f(x^2)=\frac{f(x)}{x} \implies f(x)=\frac{1}{x},$$as desired. $\blacksquare$

calvin so hawt WHY HAS CJ NO T POSTED ON THIS YET oops offtopic- anyways attached overleaf cuz looks better

The solution is $f(x) = \dfrac{1}{x}$. $P(1,1) \implies f(1) = 1$. $P(x,1) \implies xf(x^2) + f(f(x)) = f(x) (f(f(x^2)) + 1)$. $P(1,y) \implies f(f(y)) + f(y) = f(y)(1+f(f(y^2))) \implies f(y) = yf(y^2) \implies f(y^2) = \dfrac{f(y)}{y}$. \begin{align*} P(x,1) &\implies xf(x^2) + f(f(x)) = f(x) (f(f(x^2)) + 1) \\ &\implies f(x) + f(f(x)) = f(x)\cdot f(f(x^2)) + f(x) \\ &\implies f(f(x^2)) = \dfrac{f(f(x))}{f(x)} .\end{align*} Now, $f(f(x^2)) = \dfrac{f(f(x))}{f(x)} = f(f(x)^2)$. \begin{align*} P(x,y) &\implies xf(x^2) f(f(y)) + f(yf(x)) = f(xy)[f(f(x^2)) + f(f(y^2))] \\ &\implies f(x)f(f(y)) + f(yf(x)) = f(xy) [f(f(x)^2) + f(f(y)^2)] .\end{align*} \begin{align*} P \left( x,\frac{1}{x}\right) &\implies f(x)f\left(f\left(\frac{1}{x}\right)\right) + f\left(\frac{f(x)}{x}\right) = f(1)\left[\frac{f(f(x))}{f(x)} + \frac{f\left(f\left(\frac{1}{x}\right)\right)}{f\left(\frac{1}{x}\right)}\right] \\ &\implies f(x) \cdot f\left(f\left(\frac{1}{x}\right)\right) = f\left(f\left(\frac{1}{x^2}\right)\right) \\ &\implies f(x) = \dfrac{f\left(f\left((\frac{1}{x})^2\right)\right)}{f\left(f\left(\frac{1}{x}\right)\right)} = \dfrac{f\left(f\left(\frac{1}{x}\right)\right)}{f\left(\frac{1}{x}\cdot f\left(f\left(\frac{1}{x}\right)\right)\right)} \\ &\implies f\left(\frac{1}{x}\right) = \dfrac{1}{f(x)} .\end{align*} \begin{align*} P(x,x) &\implies xf(x^2)f(f(x)) + f(xf(x)) = f(x^2) \cdot 2f(f(x^2)) \\ &\implies f(x) f(f(x)) + f(xf(x)) = \dfrac{2f(x)}{x}\cdot \dfrac{f(f(x))}{f(x)} =\dfrac{2f(f(x))}{x} \\ &\implies \dfrac{f(x)}{2}+ \dfrac{f(xf(x))}{2f(f(x))} = \dfrac{1}{x}. \qquad\ (\clubsuit) \end{align*} Now change $x \rightarrow \frac{1}{x}$ to get, \begin{align*} &\dfrac{f\left(\frac{1}{x}\right)}{2} + \dfrac{f\left(\frac{1}{x}\cdot f\left(\frac{1}{x}\right)\right)}{2f\left(f\left(\frac{1}{x}\right)\right)} = x \\ \implies &\dfrac{1}{2f(x)} + \dfrac{f\left(\frac{1}{xf(x)}\right)}{2f\left(\frac{1}{f(x)}\right)} = x \\ \implies &\dfrac{1}{2f(x)} + \dfrac{f(f(x))}{2f(xf(x))} = x. \qquad\ (\heartsuit) \end{align*} From $(\clubsuit)$ and $(\heartsuit)$, we get, \[ \left(x - \dfrac{1}{2f(x)}\right)\left(\dfrac{1}{x} - \dfrac{f(x)}{2}\right) = \dfrac{1}{4}.\] This finally gives us that $f(x) = \dfrac{1}{x}$ and we are done.

Denote the assertion as $A(x,y)$. Then \begin{align*} A(1,1) &\implies f(1) = 1. \\ A(1,x) &\implies f(f(x)) = f(x) \cdot f(f(x^2)). \\ A(x,1) &\implies f(x) = xf(x^2). \\ A(x,x) &\implies xf(xf(x)) = f(f(x)) \cdot (2-xf(x)). \\ A(x^2, x^2) &\implies x^2f(x^2f(x^2)) = f(f(x^2)) \cdot (2-x^2f(x^2)). \end{align*} Noting that $xf(x) = x^2f(x^2)$ from our third substitution, we divide our last two equations to get \[x = \frac{f(f(x^2))}{f(f(x))} = \frac{f(f(x^2))}{f(x) \cdot f(f(x^2))} = \frac{1}{f(x)} \implies \boxed{f(x) = \frac 1x}. \quad \blacksquare\]

The answer is $f(x) = \frac{1}{x}.$ Let $P(x,y)$ denote the given assertion. First, we find $f(1).$ From $P(1,1),$ we get that $$f(1)f(f(1))+f(f(1)) = f(1)f(f(1))+f(1)f(f(1)),$$so $$f(f(1))(f(1)-1) = 0.$$Since $f(x) > 0,$ we have that $f(1) = 1.$ Claim: $f(y)f(f(y^{2})) = f(f(y)).$ From $P(1,y),$ we get $$f(f(y))+f(y) = f(y)+f(y)f(f(y^{2})),$$so $$f(f(y)) = f(f(y^{2}))f(y).$$We denote this by equation $1.$ Claim: $f(x) = xf(x^{2}).$ $P(x,1)$ gives $$xf(x^{2})+f(f(x)) = f(x)f(f(x^{2}))+f(x).$$By equation $1,$ this becomes $$xf(x^{2})+f(x)f(f(x^{2})) = f(x)f(f(x^{2}))+f(x),$$or $xf(x^{2}) = f(x).$ We will denote this as equation $2.$ Claim: $f$ is injective. Let $f(x)=f(z).$ Then, notice that, if we rewrite the assertions $P(x,y)$ and $P(z,y)$ as $$f(f(y))f(x)+f(yf(x)) = f(xy)\left(\frac{f(f(x))}{f(x)}+f(f(y^{2}))\right),$$and similarly for $z$ (using equations $1$ and $2$), the left hand sides are the same for $x$ and $z.$ Thus, we get that $$f(xy)\left(\frac{f(f(x))}{f(x)}+f(f(y^{2}))\right) = f(zy)\left(\frac{f(f(z))}{f(z)}+f(f(y^{2}))\right).$$Since the expressions inside the parenthesis are the same, we get that $f(xy) = f(zy).$ we will let this be equation $3.$ By two applications of equation $3,$ $$f(x^{2}y) = f(xyz) = f(yz^{2}).$$So, now, using equation $3$ again for $y = \frac{1}{y},$ $x=x^{2},$ and $z=z^{2},$ we get that $f(x^{2}) = f(z^{2}).$ Now, by equation 2, $$\frac{f(x)}{x} = f(x^{2}) = f(z^{2}) = \frac{f(z)}{z}.$$So, we find that $x=z,$ proving the claim. Now, plugging in $x = f(y)$ into equation $2$, we get that $$f(f(y)) = f(y)f(f(y)^{2}),$$but by equation $1,$ we know that $$f(f(y)) = f(y)f(f(y^{2})),$$so $f(f(y^{2})) = f(f(y)^{2}).$ By injectivity, we get that $f(y^{2}) = f(y)^{2}.$ Now, again by equation $2,$ we have that $$f(f(y^{2})) = \frac{f(y)}{y} = f(y)^{2},$$so $f(y) = \frac{1}{y},$ so we are done.

Let $P(x, y)$ denote assertion. $P(1, 1)$ \[f(f(1)f(1)=f(f(1)))\]\[f(1)=1\]$P(1, y)$ \[f(f(y))=f(y)f(f(y^2))\]$P(x, 1)$ \[xf(x^2)+f(f(x))=f(x)(f(f(x^2))+1)\]\[xf(x^2)+f(f(x))=f(x)(\frac{f(f(x))}{f(x)}+1)\]\[xf(x^2)=f(x)\]I will now prove that $f$ is injective. Suppose $f(a)=f(b)$ and $a\neq b$, $P(a, y)-P(b, y)$ \[f(ay)=f(by)\]Thus $f(a^2)=f(b^2)$ so we get $af(a^2)=bf(b^2)$ and thus $a=b$. Thus $f$ is injective. We also have that $f(x)f(f(x)^2)=f(x)f(f(x^2))$ which implies $f(x^2)=f(x)^2$ so we get $f(x)=\frac{1}{x}$ as the solution.

Amazing FE! My solution is much more complicated than those on the thread. The only solution is $f(x) = \frac{1}{x}$, which clearly works. Now we shall prove this is the only solution. Let $P(x,y)$ denote the given assertion. Claim 1: $f(1) = 1.$ Proof: Use $P(1,1)$ to get $$2 f(f(1)) = 2 f(1) f(f(1)).$$Since $f(f(1)) \ne 0,$ we conclude that $f(1) = 1.$ Claim 2: $x f(x^2) = f(x).$ Proof: Use $P(x,1) - P(1,x)$ to get $$x f(x^2) f(f(1)) + f(f(x)) - f(1) f(f(x)) - f(x f(1)) = x f(x^2) f(1) + f(f(x) - f(f(x)) - f(x) = x f(x^2) - f(x) = 0,$$implying the claim. Claim 3: $f(f(x^2)) = f(f(x)) f\left(\frac{1}{x}\right)$. Proof: Use $P\left(\frac{1}{x}, x\right)$ to get $$\frac{1}{x} f\left(\frac{1}{x^2}\right) f(f(x)) + f\left(\frac{f\left(\frac{1}{x}\right)}{\frac{1}{x}}\right) = f(1) \left(f(f(x^2)) + f\left(f\left(\frac{1}{x^2}\right)\right)\right).$$By Claim 2, we have $\frac{f\left(\frac{1}{x}\right)}{\frac{1}{x}} = f\left(\frac{1}{x^2}\right),$ so plugging that in and using $f(1) = 1$ gives $$f\left(\frac{1}{x}\right) f(f(x)) + f\left(f\left(\frac{1}{x^2}\right)\right) = f(f(x^2)) + f\left(f\left(\frac{1}{x^2}\right)\right).$$Several terms cancel out, leaving us with $$f(f(x^2)) = f(f(x)) f\left(\frac{1}{x}\right),$$as claimed. Claim 4: $f(x) f\left(\frac{1}{x}\right) = 1.$ Proof: Use $P(x,1)$ along with Claim 2 to get $$f(x) + f(f(x)) = f(x)(1+f(f(x^2)).$$Expanding the right-hand side and subtracting $f(x)$ from both sides, we get $$f(f(x)) = f(x) f(f(x^2)).$$But by Claim 3, we know that $f(f(x^2)) = f(f(x)) f\left(\frac{1}{x}\right).$ This forces $f(x) f\left(\frac{1}{x}\right) = 1,$ as claimed. Claim 5: $x f(f(x)) = f(f(x^2)).$ Proof: Use $P(x,x)$ to get $$x f(x^2) f(f(x)) + f(x f(x)) = f(x^2) f(f(x^2)) + f(x^2) f(f(x^2)).$$Then use $P\left(\frac{1}{x}, \frac{1}{x}\right)$ and repeatedly use Claim 4 to get $$\frac{1}{x f(x^2) f(f(x))} + \frac{1}{f(x f(x))} = \frac{1}{f(x^2) f(f(x^2))} + \frac{1}{f(x^2) f(f(x^2))}.$$Say by Vieta, this implies that the terms on the left-hand side of $P(x,x)$ are a permutation of the terms on its right-hand side. However, both terms on the right-hand side are equal to $f(x^2) f(f(x^2)),$ which gives us $x f(x^2) f(f(x)) = f(x^2) f(f(x^2)),$ and dividing both sides by $f(x^2)$ gives us our claim. To finish, we use Claims 3 and 5 to get $$f(f(x^2)) = x f(f(x)) = f(f(x)) f\left(\frac{1}{x}\right).$$This implies that $f\left(\frac{1}{x}\right) = x,$ and therefore $f(x) = \frac{1}{x}$ for all $x,$ as desired.

The answer is $f(x)=\frac 1x$. Let $P(x,y)$ denote the assertion. $P(1,1)$ gives that $f(1)=1$. Then, $P(x,1)$ and $P(1,x)$ give \[xf(x^2)+f(f(x))=f(x)(1+f(f(x^2)))=f(f(x))+f(x)\]so $xf(x^2)=f(x)$ and $f(f(x))=f(f(x^2))f(x)$. Thus \[f(f(x^2))=\frac{f(f(x))}{f(x)}=f(f(x)^2),\]so if $f$ is injective, then $f(x)^2=f(x^2)=\frac{f(x)}{x}$ as we want. Thus the following suffices. Claim: $f$ is injective. Proof. Suppose $f(x_1)=f(x_2)$. Then, $P(x_1,y)$ and $P(x_2,y)$ give $f(x_1y)=f(x_2y)$ so $f(x_1^2)=f(x_1x_2)=f(x_2^2)$. However, \[x_1f(x_1^2)=f(x_1)=f(x_2)=x_2f(x_2^2)\]implying $x_1=x_2$. $\blacksquare$

The answer is $$f(x)=\frac{1}{x}$$for all $x \in \mathbb{R}$. It's easy to see that these functions satisfy the given equation. Now, we shall show that these are the only ones. First start by letting $x=y=1$ to receive $$f(1)=1$$Now, look at $P(x,1)$ and $P(1,x)$ to see that we have $$f(x)=xf(x^2)$$and $$f(f(x))=f(x)f(f(x^2))$$Now, that we have these look at $P(\frac{1}{x},\frac{1}{x})$. We them have (using the previous two equations to simplify) \begin{align*} \frac{1}{f(x)f(f(x))} + \frac{1}{f(xf(x))} &= \frac{1}{f(x^2)f(f(x))} + \frac{1}{f(x^2)f(f(x))}\\ f^2(x)f(x^2)\left(f(xf(x)) + f(x)f^2(x)\right) &= f(xf(x))f^2(x) \left(f(x)^2 + f(x)^2\right)\\ f(x^2)f(xf(x)) + f(x^2)f(x)f^2(x) &= f(xf(x))f(x^2) + f(x)^2f(xf(x))\\ f(xf(x)) &= \frac{f(x^2)f(f(x))}{f(x)}\\ f(xf(x)) &= \frac{f(f(x))}{x} \end{align*}Now, plugging this back into $P(x,x)$ we easily have that \begin{align*} f(x)f(f(x)) + f(xf(x)) &= f(x^2)2f(f(x^2))\\ f(x)f(f(x)) + \frac{f(f(x))}{x} &= \frac{f(x)}{x} \frac{2f(f(x))}{f(x)}\\ f(x)+ \frac{1}{x} &= \frac{2}{x}\\ f(x) &= \frac{1}{x} \end{align*}which was the required conclusion. Thus, we directly have that the only functions satisfying the original equation are in fact of the claimed type.

The answer is $f(x) = \tfrac{1}{x}$, which can easily be checked to work. Firstly, taking $P(1,1)$ gives $f(1) = 1$. Claim: We have $f(x) = xf(x^2)$. Proof: Taking $P(1,x)$ gives us $f(f(x)) = f(x) f(f(x^2))$ while taking $P(x,1)$ gives us $xf(x^2) + f(f(x)) = f(x) ( f(f(x^2)) + 1)$. Subtracting the two gives us the desired claim. Claim: We have $f(f(x)) = f(x)f(f(x^2))$. Proof: This is $P(1, f(x))$. Claim: We have $f( \tfrac{1}{x} ) f(x) = 1$. Proof: Taking $P( \tfrac{1}{x} ,x)$ gives us (after simplifying the first $xf(x^2)$ in the given FE with the first claim) \[f( \tfrac{1}{x} ) f(f(x)) = f(f(x^2)).\]Substituting with the second claim gives us the desired claim. Now, compare $P(x,y)$ and $P( \tfrac{1}{x}, \tfrac{1}{y})$. Each of four terms in the second expression is a reciprocal of the corresponding term in the first expression, so we have a system of equations like $a+b = c+d$ and $\tfrac{1}{a} + \tfrac{1}{b} = \tfrac{1}{c} + \tfrac{1}{d}$. Due to, say, Vieta's formulas, this implies that $(a,b)$ and $(c,d)$ are permutations of each other. In particular, the special case of $x=y$ tells us that \[xf(x^2) f(f(x)) = f(x^2) f(f(x^2)).\]Dividing both sides by the second claim finishes.

We claim that the only solution is $f\equiv x^{-1}$, which obviously works; we now prove that it is the only one. Let $P(x,y)$ denote the assertion. Claim: $f(1)=1$. Proof: By $P(1,1)$, we have $f(1)f(f(1))+f(f(1)) = 2f(1)f(f(1))$, so $f(1)=1$. Claim: $f(x)=xf(x^2)$ and $f(x)f(f(y)) + f(yf(x)) = f(y)f(f(x)) + f(xf(y))$ for all positive reals $x$ and $y$. Proof: Comparing $P(x,y)$ and $P(y,x)$ gives $xf(x^2)f(f(y)) + f(yf(x)) = yf(y^2)f(f(x)) + f(xf(y)) $. Note that using this equation, the second part of our claim follows directly from the first. Plugging in $y=1$, we have $xf(x^2)+f(f(x)) = f(f(x))+f(x)$, which implies $f(x)=xf(x^2)$. For future reference, let $Q(x,y)$ denote the assertion $f(x)f(f(y)) + f(yf(x)) = f(y)f(f(x)) + f(xf(y))$. Claim: $f(f(x)^2) = f(f(x^2))$ for all positive reals $x$. Proof: From our previous claim, we have $f(f(x)) = f(x)f(f(x)^2)$ by substituting $f(x)$ for $x$. On the other hand, from $P(1,x)$, we obtain $f(f(x)) + f(x) = f(x)(1+f(f(x^2)))$, which simplifies to $f(f(x)) = f(x)f(f(x^2))$. Thus, $f(x)f(f(x)^2) = f(x)f(f(x^2))$, which implies $f(f(x)^2) = f(f(x^2))$. Claim: $f$ is injective. Proof: Suppose $a$ and $b$ are positive reals satisfying $f(a)=f(b)$; we show that $a=b$. Let $N$ denote the common value of $f(a)$ and $f(b)$. By $Q(a,b)$, we have $Nf(N) + f(bN) = Nf(N) + f(aN)$, or $f(aN)=f(bN)$. Let $M$ denote the common value of $f(aN)$ and $f(bN)$. Then, $P(a,a)$ and $P(b,b)$ give us that $Nf(N) + M = 2f(a^2)f(f(a^2)) = 2f(b^2)f(f(b^2))$. However, since $f(f(a^2)) = f(f(a)^2) = f(f(b)^2) = f(f(b^2))$, we must have $f(a^2)=f(b^2)$. Thus, $a = \tfrac{f(a)}{f(a^2)} = \tfrac{f(b)}{f(b^2)} = b$, as desired. Claim: $f(x) = \tfrac{1}{x}$ for all positive reals $x$. Proof: From $f(f(x)^2) = f(f(x^2))$ and the injectivity of $f$, we deduce that $f(x)^2 = f(x^2)$. Hence, we have $f(x) = xf(x)^2$, or $f(x) = \tfrac{1}{x}$, which proves the claim and completes our proof. $\blacksquare$