Let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ gives $f(0)=0$, now divide $x$ on both sides of the F.E. and apply symetry and the multiply by $xy$ on both sides to get: $$x^2yf(y)+y^2f(x)^2=y^2xf(x)+x^2f(y)^2 \; \forall x,y \in \mathbb R$$Call this $Q(x,y)$, by $Q(x,1)$ we get $f(x)^2-xf(x)+x^2f(1)-x^2f(1)^2=0$, and now from the quadratic formula we get that: $$f(x)=\frac{x \pm \sqrt{x^2(4f(1)^2-4f(1)+1)}}{2}=\frac{x \pm (x(2f(1)-1))}{2} \implies f(x)=xf(1) \; \text{or} \; x(1-f(1))$$Suppose that $f(a)=af(1)$ and $f(b)=b(1-f(1))$ exist for $a,b \ne 0$, by $P(a,b)$ we get: $$af(ab)=a^2b(1-f(1))+a^2bf(1)^2 \implies f(ab)=ab(f(1)^2-f(1)+1) \implies f(1)=1 \; \text{or} \; 0$$If $f(1)=0$ we get $f(a)=0$ so by $P(a,x)$ we get $f(af(x))=af(x)$, now if $a=1$ if the only zero we have $f(x)=x$ for all $x \ne 1$ and $f(1)=0$ which fails because in this case we have $f \left(\frac{1}{\sqrt{2}} \right)=\frac{1}{\sqrt{2}}$ and by setting $x=y=\frac{1}{\sqrt{2}}$ we get that $f(1)=1$, if $a$ takes more values than just $1$ then we have $f(a^kb)=a^kb$ for $k \ge 0$ integer, note that the F.E. now became $f(xf(y)+yf(x))=xf(y)+yf(x)$ for all $x,y \in \mathbb R$. Therefore $P(x,b)$ gives $f(b(x+f(x)))=b(x+f(x))$ so $f(2bf(x))=2bf(x)$, denote $A$ the set of all zeros of $f$ and $B$ the set of all identity points on $f$, its clear that $A \cup B=\mathbb R$, we have that if $a,b \in B$ then $2ab \in B$, we have $0,1 \in A$ and also note that if $a \in A$ and $b \in B$ then $ab \in B$, suppose for some $a,a' \in A$ we have $\frac{a'}{a} \in B$ then $a' \in B$ contradiction!. Now notice that it is clear that there is infinite elements in $B$, and also that with the previous claim we proved that if $x \in A$ then $x^{-1} \in A$ as well, now suppose $A$ is finite, then since we have $a^kb \in B$ for any $k$ integer and $a \in A$ and $b \in B$ we must have that $\frac{a'}{a^k} \in A$ but the only case were we avoid infinitely elements is when only $1,-1 \in A$ which gives $f(1)=f(-1)=0$ and $f(x)=x$ for all $x \ne 1,-1$ which also fails because once again $\frac{1}{\sqrt{2}}$ is an identity point and by $x=y=\frac{1}{\sqrt{2}}$ we get $f(1)=1$. Now if $A$ was not finite then it must have an element distinct from $1,-1$, note that we from this and the previous observation, have that $A$ is also closed under multiplication and note that if for some $a \in A$ we have $\frac{a}{\sqrt{2}} \in B$ then we get $a^2 \in B$ so we must have $\frac{a}{\sqrt{2}} \in A$ as well, now if $\sqrt{a} \in B$ for some $a \in A$ then we have $2a \in B$ but then $2a^2 \in B$, however we can also get $2a^2 \in A$ frommthe previous claims, therefore $\sqrt{a} \in A$ for any $a \in A$, from this we have that $2^{\pm 2^k} \in A$ for any $k$ integer. Now the thing is that for $k \ne 0$ we get that if both $k, \frac{a}{k} \in B$ for $a \in A$ then $2a \in B$ which cannot happen therefore either $k \in A$ or $\frac{a}{k} \in A$, but since $a^{-1} \in A$, from either case we get that $k \in A$, if we choose $k=b$ then we would get a contradiction!. If $f(1)=1$ then we have $f(a)=a$ and $f(b)=0$, but $P(b,1)$ gives $f(b)=b$ so $b=0$ which is a contradiction!. Therefore from the split cases we get no solutions. Now if such $a,b$ didn't exist, in either case we have $f$ linear, trivially if $f$ is constant then $f$ is $0$ so assume $c \ne 0$, then replacing on the original F.E. we get $c^2x^2y+c^2x^2y=cx^2y+c^2x^2y$ therefore $c=1$, so $f(x)=x$. Therefore, all the functions that satisfy the following F.E. are $f(x)=x, f(x)=0$, thus we are done .

By using symetrry $xf(y)+y\frac{f(x)^2}{x}=yf(x)+x\frac{f(y)^2}{y}$ $\implies \frac{f(y)}{y}+(\frac{f(x)}{x})^2=\frac{f(x)}{x}+\frac{f(y)^2}{y^2}$ . for any non zero $x,y$ This means if $S$ is the set of all values $\frac{f(x)}{x}$ takes then $|B| \leq 2$ and if it has two elements the sum of those 2 is $2$ . Putting $x=0$ , its eays to see $f(0)=0$ if $|B|=1$ the prob is trivial . SO we assume $|B|=2$ Define $g(x)=\frac{f(x)}{x}$ for all non zero $x$ Then[with bit of observation is easy to see]. Case 1: $g(x)$ is never $0$ $x^2yg(xy(g(y)+g(x))=x^2y(g(y)+g(x)^2)$ so $g(xy(g(y)+g(x))(g(y)+g(x))=g(y)+g(x)^2$ The main thing is that then $g(y)+g(x)$ is never $0$. Let $g(y)$ be the global maximum of $g$ , now if there exist $g(x) \neq g(y)$ then $g(x)+g(y)=1$ and we have a contradiction. This would $g$ is constant and then prob is trivial. Case 2: $B=\{0,1\}$. Claim: if $f(x)=0$ then $x=0$. Proof: assume the contradiction observe $xf(xf(y))=x^2f(y)$ This attains $f$ attains all non zero values then $f(x)=x$. hence done.

Let $P(x,y)$ denote the given assertion. $P(0,1): f(0)^2 = 0\implies f(0) = 0$. $P(x,x): xf(2xf(x)) = x(xf(x) + f(x)^2)$, so $f(2xf(x)) = xf(x) + f(x)^2$ for $x\ne 0$ and it's obviously true for $x = 0$ also. Let $g(x) = f(x) - x$. We have $x(xf(y) + yf(x)) + xg(xf(y) + yf(x)) = x^2 f(y) + y f(x)^2$. $ xg(xf(y) + yf(x)) = yf(x)^2 = yg(x)^2 + xy g(x)$ $yg(xf(y) + yf(x)) = xg(y)^2 + xyg(y)$ $\frac{x^2 f(y) + yf(x)^2}{x} = \frac{y^2 f(x) + x f(y)^2}{y}$. So $x^2 y f(y) + y^2 f(x)^2 = x y^2 f(x) + x^2 f(y)^2$. $x^2(yf(y) - f(y)^2) = y^2(xf(x) - f(x)^2)$, so $\frac{xf(x) - f(x)^2}{x^2}$ is constant, so $xf(x) - f(x)^2 = cx^2$ for some constant $c$. Let $f(x) = xg(x)$. We have $x^2 g(x) - x^2 g(x)^2 = cx^2$, so $g(x) - g(x)^2 = c$ for some constant $c$. Now, we have $c = f(1) - f(1)^2 = g(1) - g(1)^2$, so $g(x) - g(x)^2 = g(1) - g(1)^2$, implying $g(x) \in \{g(1), 1 - g(1)\} = \{f(1), 1 - f(1)\}$ for all $x$. Hence $f(x) \in \{x f(1), x(1 - f(1))\}$. $P(1,1): f(2f(1)) = f(1)^2 + f(1)$, so either $2f(1)^2 = f(1)^2 + f(1) \implies f(1) \in \{0,1\}$ or $2f(1)(1 - f(1)) = f(1)^2+ f(1) \implies f(1) \in \left \{ 1, \frac 13 \right \}$, so $f(1)$ is either $0, 1, $ or $\frac 13$. Case 1: $f(1) \in \{0,1\}$ Then $f(x) \in \{0,x\}$ for all $x$. Claim: If $x,y$ are fixed points, then $2xy$ is also. Proof: Trivial by $P(x,y)$ and dividing both sides by $x$ (and the result is obvious for $x = 0$). $\square$ Claim: If $x$ is a fixed point and $y$ isn't, then $xy$ is a fixed point. Proof: Since $f(y) = 0$, $P(x,y)$ gives that $xf(xy) = x^2 y$, so $f(xy) = xy$ or $x = 0$, but clearly $f(xy) = xy$ if $x =0$. $\square$ Claim: If $x$ is a fixed point, then $4x$ is. Proof: If $2$ wasn't a fixed point, then $x$ fixed point implies $2x$ fixed point, so $4x$ also is a fixed point. Otherwise, if $2$ is a fixed point, then $2\cdot x \cdot 2 = 4x$ is a fixed point. $\square$ Now notice that if $x$ is a fixed point, then $2x^2$ also is. Therefore, if $x,y$ are fixed points, then $2(2xy)^2 = 8x^2y^2$ is a fixed point. However, if $x$ is a fixed point and $y$ isn't, then $xy$ is a fixed point, so $2(xy)^2 = 2x^2y^2$ is a fixed point, meaning that $8x^2 y^2$ also is. Hence $8x^2y^2$ is a fixed point whenever at least one of $x,y$ is a fixed point. Suppose $f$ was not the identity. Subcase 1.1: There exists positive $k$ that isn't a fixed point. We show that all positive reals are not fixed points. Suppose that there was one positive real $y$ with $f(y) = y$. Then consider positive $x$ with $x^2 y^2 = \frac k8$, so $8x^2y^2 = k$ must be a fixed point, absurd. Hence all positive reals are not fixed point and thus are roots of $f$. Now, if there was a negative fixed point $k$, then $2k^2$ is also a fixed point, but that is positive, contradiction. Subcase 1.2: All positive real numbers are fixed points. Claim: There is a negative fixed point. Proof: Suppose otherwise. Then $-1$ isn't a fixed point, so if you consider $x$ positive, we have $x \cdot -1$ is a fixed point, so there must be a negative fixed point. $\square$ Now let $k$ be this negative fixed point. Notice that $2kx$ is a fixed point for all positive $x$, but this spans all negative numbers, meaning all negative reals are also fixed points. Since $f(0) = 0$, $f$ must be the identity. Case 2: $f(1) = \frac 13$ Then for each $x$, $f(x)$ is either $\frac x3$ or $\frac{2x}{3}$. If $f(x) = \frac{2x}{3}$ for some $x$, we have $P(x,x): f(2xf(x)) = xf(x) + f(x)^2$, so $f \left( \frac{4x^2}{3} \right) = \frac{2x^2}{3} + \frac{4x^2}{9} = \frac{10x^2}{9}$, which is clearly impossible for $x \ne 0$. Hence $f(x) = \frac x3$ for all $x$, which doesn't work.

Let $P(x, y)$ denote the given assertion. It is easy to see that $f(0) = 0$. From now on, we will consider only non-zero variables. By symmetry, we have \[xf(y) + \frac{y}{x} f(x)^2 = yf(x) + \frac{x}{y} f(y)^2 \iff \left(\frac{f(x)}{x} - \frac{f(y)}{y} \right)\left(\frac{f(x)}{x} + \frac{f(y)}{y} -1 \right)=0\]Hence, $f(x) \in \{cx, (1-c)x\}$ for each $x$, where $c = f(1)$. From, \begin{align} P(1, 1): \thinspace f(2c) = c+c^2, \end{align}it follows that $c \in \{1/3, 1, 0\}$, by considering the possible values of $f(2c)$. Now, we solve these three cases separately. Case (i): Suppose $c=1/3$. Then, Eq(1) gives $f(2/3) = 4/9$. But, \begin{align*} P(1, 2/3)&: f(2/3) = f(4/9 + (2/3)(1/3)) = 4/9 + 2/3(1/3)^2 = 14/27 \end{align*}which is absurd. Case (ii): Suppose $c = 1$. If there exists $a$ such that $f(a) = 0$, then this leads to a contradiction since \[P(a, 1): \thinspace af(a)=a^2 \implies a=0.\]Hence, $f(x) = x$ for all $x$. Case (iii): Suppose $c=0$. Recall that $f(x) \in \{0, x\}$ for each $x$. Let $\mathcal{F}$ denote the set of fixed points of $f$. It is easy to see that \begin{align} \setcounter{equation}{1} &\text{if }x, y \in \mathcal{F}, \text{ then } 2xy \in \mathcal{F}, \quad \text{and}\\ &\text{if } x \in \mathcal{F} \text{ and } y\notin \mathcal{F}, \text{ then } xy \in \mathcal{F}. \end{align}Claim iii.1: If $b \in \mathcal{F}$, then $1/2, 2, 1/b, \in \mathcal{F}$ and $2b \notin \mathcal{F}$. Proof of claim: If $\frac{1}{2b} \in \mathcal{F}$, then $f(1) = f\left(2\cdot b \cdot \frac{1}{2b} \right) = 1$, which is absurd. Thus, $\frac{1}{2b} \notin \mathcal{F}$. Similarly, \begin{align*} b \in \mathcal{F} \text{ and } \frac{1}{2b} \notin \mathcal{F} &\implies 1/2 \in \mathcal{F} \\ 1/2 \in \mathcal{F}, 1 \notin \mathcal{F} \text{ and } (3) &\implies 2 \in \mathcal{F}.\\ \frac{1}{2b} \notin \mathcal{F} \text{ and } 2 \in \mathcal{F} &\implies \frac{1}{b} \in \mathcal{F} \\ \frac{1}{2b} \notin \mathcal{F}, 1 \notin \mathcal{F} \text{ and } (3) &\implies 2b \notin \mathcal{F}. \end{align*} Claim iii.2: If $b \in \mathcal{F}$, then $b^2 \notin \mathcal{F}$. Proof of claim: If $b^2 \in \mathcal{F}$, then $2b^2 \notin \mathcal{F}$ by Claim iii.1. But since $b \in \mathcal{F}$, (2) implies that $2b^2 \in \mathcal{F}$, hence a contradiction. Now, assume for the sake of contradiction, that $f(a) = a$ for some $a\neq 0$. Then, $2a^2 \in \mathcal{F}$ by (2). The contrapositive of Claim iii.2 shows that $\sqrt{2}a \notin \mathcal{F}$. From this and $1/2 \in \mathcal{F}$ and (3), it follows that $\frac{a}{\sqrt{2}} \in \mathcal{F}$. By (2), $a^2 \in \mathcal{F}$ which contradicts Claim iii.2. Hence, $f(x)= 0$ for all $x$ in this case.

Was I lucky (is my solution a fakesolve) or is this problem actually trivial Let assertion \( P(x, y) = x f(x f(y) + y f(x)) = x^2 f(y) + y f(x)^2 \). \( P(0, x) \) gives \( f(0) = 0 \). \( P(y, x) = y f(x f(y) + y f(x)) = y^2 f(x) + x f(y)^2 \). So we have \[ f(x f(y) + y f(x)) = \frac{x^2 f(y) + y f(x)^2}{x} = \frac{y^2 f(x) + x f(y)^2}{y}. \]On further algebraic simplification, we get \[ \frac{f(x)(x - f(x))}{x^2} = \frac{f(y)(y - f(y))}{y^2} = \text{constant} = k \quad (\text{say}). \] Applying the quadratic formula, we get \[ f(x) = x \left( \frac{1 \pm \sqrt{1 - 4k^2}}{2} \right). \]Since \( \frac{1 \pm \sqrt{1 - 4k^2}}{2} \) is independent of \( x \) and only depends on the constant \( k \), we can assume it to be another constant \( p \). So, \( f(x) \) is of the form \( px \). Substituting in assertion, we get \[ x^2 p^2 y = x^2 p y \quad \text{for all } x, y \in \mathbb{R}. \]Hence, \( p^2 = p \), which means \( p = 1 \) or \( p = 0 \). Hence, \( f(x) = 0 \) or \( f(x) = x \) are the only solutions. \(\boxed{\text{QED}}\)