For $y=(a+f(a))f(b)$ the $RHS$ became symmetriv for $a,b$ $f([(f(a)+a)f(b)+f([f(a)+a]f(b))]f(x))=f(f(a)f(b)f(x)+af(b)f(x)+bf(a)f(x)+f(ab)f(x))$ So we must have: $xf([a+f(a)]f(b))+f([a+f(a)]f(b)x)=xf([b+f(b)]f(a))+f([b+f(b)]f(a)x)$ For $x=1$ and from the first this one became: $2f([a+f(a)]f(b))=2f([b+f(b)]f(a))\Rightarrow 2bf(a)+2f(ab)=2af(b)+2f(ab)\Rightarrow bf(a)=af(b)$ So we have that: $\frac{f(a)}{a}=\frac{f(b)}{b}=\frac{f(1)}{1}=f(1)\Rightarrow f(a)=ac$ for all $a\neq 0$ Taking at the first $x,y$ big we get: $c^2xy+c^3x=cxy+cxy\Rightarrow c^3+c^2-2c=0\Rightarrow c(c+2)(c-1)=0\Rightarrow c=0,1,-2$ If $f(0)\neq 0$ then $P(x,0)$ gives $c^2xf(0)=(x+1)f(0)$ contradiction So the sollution are:$f(x)=x,0,-2x$

Let $P(x,y)$ be the assertion in the hypothesis. If $f(1)=0$, $P(1,y)$ gives $f\equiv 0$, which works. If $f(1)\neq 0$, $P(x,1)$ gives that $f$ is injective. Also, consider $g(x)=f(x)+x$. From $ P(1,y)$ we get $$f(g(y)f(1))=2f(y),$$and since $f$ in injective, $g$ must be as well. Finally, adding $yf(x)+f(x)f(y)$ to both sides of $P(x,y)$ gives $$g(yf(x)+f(x)f(y))=xf(y)+yf(x)+f(xy)+f(x)f(y).$$Notice that the RHS is symmetric in $x$ and $y$, so swapping the variables and using the injectivity of $g$ we get $xf(y)=yf(x)$, from which $f(x)=cx$ and finally obtaining the last two solutions, $f(x)=x$ and $f(x)=-2x$.

The only solutions are $\boxed{f(x) = -2x}, \boxed{f \equiv 0}, \boxed{f(x) = x}$, which work. Now we prove they are the only solutions. Let $g(x) = f(x) + x$. The equation becomes $f(f(x) g(y)) = xg(y) + g(xy) - 2xy$, so $g((g(x) - x) g(y)) - (g(x) - x) g(y) = xg(y) + g(xy) - 2xy$, which simplifies to \[ g((g(x) - x) g(y)) - g(x)g(y) = g(xy) - 2xy\]Let $P(x,y)$ denote this assertion. Now, if $g(a) = g(b)$, then $P(1, a)$ compared with $P(1, b)$ gives that $a = b$. Now comparing $P(x,y)$ and $P(y,x)$ gives $g((g(x) - x) g(y)) = g((g(y) - y) g(x))$, so \[(g(x) - x) g(y) = (g(y) - y) g(x),\]implying that $xg(y) = yg(x)$ for all reals $x,y$. This implies $g(x) = xg(1)$ for all reals $x$ by setting $y = 1$. Let $g(1) = c$ so we have $g(x) = cx$. $P(1,1)$ gives that $g(c(c-1)) - c^2 = c - 2$, so $c^2(c-1) = c^2 + c - 2 = (c+2)(c-1)$. Hence $c = 1$ or $c^2 = c + 2\implies c \in \{-1, 2\}$. Therefore since $f(x) = g(x) - x$, $f(x) = -2x$, $f(x) = 0$, and $f(x) = x$ are the only solutions.

Finally allowed to post Iran TST? This is peak reals to reals FE. Let $P(x,y)$ denote the given FE. Note that $$\boxed{f(x)=0 \ \ \forall x \in \mathbb{R}}$$is the only constant solution, so let us look for non-constant ones. Claim 1: $f$ is injective. Proof: If $f(1)=0$, then $P(1,y)$ gives $2f(y)=f(0)$ for all $y$, so $f$ is constant, which is the case we have discarded. Thus $f(1) \neq 0$. But then $P(x,1)$ gives $f(f(x)(y+f(y)))=xf(1)+f(x)$, which immediately implies injectivity of $f$. $\blacksquare$ Claim 2: $f(y)+y$ is an injective function. Proof: $P(1,y)$ implies $f(f(1)(y+f(y)))=2f(y)$, and the claim thus follows from injectivity of $f$. $\blacksquare$ Now the magic: compare $P(x,y)$ and $P(y,x)$ to get $$f(yf(x)+f(x)f(y))+yf(x)+f(x)f(y)=xf(y)+yf(x)+f(xy)+f(x)f(y)=f(xf(y)+f(x)f(y))+xf(y)+f(x)f(y).$$Hence Claim 2 implies $yf(x)+f(x)f(y)=xf(y)+f(x)f(y)$, or $yf(x)=xf(y)$ for all reals $x,y$. Putting $y=1$ gives $f(x)=cx$ for some constant $c$, and putting this solution back in $P(x,y)$, we see that the only non-zero $c$ that work are $1$ and $-2$, so the only other solutions are: $$\boxed{f(x)=x \ \ \forall x \in \mathbb{R}}$$$$\boxed{f(x)=-2x \ \ \forall x \in \mathbb{R}}$$

This was kinda hard, as in non-standard. First time I ever saw such a goofy ahh substitution actually work. Now my solution looks a thousand times worse compared to the ones above. The answers are $f(x)=0$ for all $x \in \mathbb{R}$ , $f(x)=x$ for all $x \in \mathbb{R}$ and $f(x)=-2x$ for all $x \in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We denote by $P(x,y)$ the assertion that $f(yf(x)+f(x)f(y))=xf(y)+f(xy)$. Let $f(1)=c$ for some real number $c$. Now, $P(1,x)$ and $P(x,1)$ gives \[f(cx+cf(x))=2f(x)\]and \[f((c+1)f(x))=cx+f(x)\]For the sake of convenience, we let $A=cx+cf(x)$ and $B=(c+1)f(x)$. Then, $P(1,A)$ gives \begin{align*} f(cA+cf(A))&=2f(A)\\ f(c^2x+c^2f(x)+2cf(x)) &= 4f(x)\\ f((c^2+2c)f(x)+c^2x) &= 4f(x) \end{align*}Similarly, $P(1,B)$ gives us \begin{align*} f(cB+cf(B)) &= 2f(B)\\ f(c(c+1)f(x)+c^2x+cf(x)) &= 2cx+2f(x)\\ f((c^2+2c)f(x)+c^2x) &= 2cx+2f(x) \end{align*}Thus, combining these results we have that \[4f(x) = f((c^2+2c)f(x)+c^2x) = 2cx+2f(x)\]Thus, $f(x)=cx$ for all real numbers $x$. Now, we can simply substitute. Consider $P(x,y)$ for any pair of real numbers $x,y\neq 0$. This gives, \begin{align*} f(yf(x)+f(x)f(y)) &= xf(y)+f(xy)\\ c^2xy + c^3xy &= 2cxy\\ c^3+c^2 -2c &= 0\\ c(c+2)(c-1) &= 0 \end{align*}Thus, $c\in \{0,1,-2\}$ which each give each of the solutions claimed above, so we are done.

Solved with the help of Orthogonal. and megarnie. We claim the only functions are $\boxed{f(x)\equiv 0}$, $\boxed{f(x)\equiv x}$, and $\boxed{f(x)\equiv-2x}$. It is easy to check that these work. Let $P(x,y)$ denote the given assertion. $P(x,1)$ gives $f(f(x)(1+f(1)))=xf(1)+f(x)$ so \[ x=\frac{f(f(x)(1+f(1)))-f(x)}{f(1)}. \]If $f(1)=0$ then $P(1,1)$ gives $f(0)=0$ and $P(1,x)$ gives $f(x)=f(0)=0$. Suppose $f(1)\neq 0$. It follows that $f$ is injective. Then $P(0,0)$ gives $f\left(f(0)^2\right)=f(0)$ so $f(0)=0$. From $P(1,x)$ we get $f(f(1)(x+f(x)))=2f(x)$ so $x\mapsto x+f(x)$ is injective. Comparing $P(x,y)$ and $P(y,x)$ yields \[ yf(x)+f(x)f(y)+f(yf(x)+f(x)f(y))=xf(y)+yf(x)+f(xy)+f(x)f(y)=xf(y)+f(x)f(y)+f(xf(y)+f(x)f(y)) \]so $yf(x)+f(x)f(y)=xf(y)+f(x)f(y)$. It follows that $\frac{f(x)}{x}$ is constant for $x\neq0$ so $f(x)\equiv cx$ for some $c$. Substituting this back into the original equation gives $c^2+c^3=2c$ so $c\in\{-2,0,1\}$, as desired. $\square$

Very difficult for P4 The only such functions are $f(x)=0$, $f(x)=x$, and $f(x)=-2x$, which can be directly verified. Let $P(x,y)$ be the given assertion. Claim: $g(x)=f(x)+x$ is injective If $g(a)=g(b)$ then comparing $P(1,a)$ and $P(1,b)$ yields $f(a)=f(b)\Rightarrow a=b$. Claim: $f(x)$ is linear Adding $yf(x)+f(x)f(y)$ to both sides we get $g(yf(x)+f(x)f(y))=xf(y)+f(xy)+yf(x)+f(x)f(y)$ Thus $g(yf(x)+f(x)f(y))$ must be symmetric in $x$ and $y$ which due to injectivity implies that $yf(x)=xf(y)$. Now let $f(x)=cx$. $P(1,1)$ gives $c^2(c+1)=2c\Rightarrow c=-2,0,1$, as desired.

Nice Problem~~ Let $P(x,y)$ be the assertion $f(yf(x)+f(x)f(y))=xf(y)+f(xy)$ if $f(1)=0$, $P(1,x)$ $\implies$ $f(x)=\frac{f(0)}2=c$. Plugging back we know $c=0$. So $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ (which indeed fits since $LHS=0=RHS$) otherwise, consider if $f(a)=f(b)$ for some $a$, $b$ $P(a,1)-P(b,1)$ $\implies$ $af(1)=bf(1)$, so $a=b$ since $f(0)\neq 0$. Therefore, $f$ is injective Define another function $g:\mathbb{R} \to \mathbb{R}$ $g(x)=f(x)+x$ $\forall x$ Notice thate $f(f(x)g(y))=xf(x)+f(xy)$ (Let $Q(x,y)$ be the assertion of this equation) consider if $g(a)=g(b)$ for some $a$, $b$ $Q(1,a)-Q(1,b)$ $\implies$ $2f(a)=2f(b)$, so $a=b$ since $f$ is injective Therefore, $g$ is also injective Rewrite the original FE with $g$ and tidy up $\implies$ $g(f(x)g(y))-g(x)g(y)=g(xy)-2xy$ Swap $x$, $y$ and comparing the asymmetrical terms. We get $g(f(x)g(y))=g(f(y)g(x))$ Since $g$ is injective, we can obtain $f(x)g(y)=f(y)g(x)$ Finally, from (*) we know $xf(y)+f(xy)=f(f(x)g(y))=f(g(x)f(y))=yf(x)+f(xy)$ So $xf(y)=yf(x)$ $\forall x,y$ namely, $\frac{f(x)}{x}$ is a constant $\forall x$ Let $f(x)=cx$ and plugging back, we get $c^2xy+c^3xy=2cxy$ $\implies$ $c^3+c^2-2c=0$ $\implies$ $c(c+2)(c-1)=0$ $\implies$ $c=0,1,-2$ If $c=0$ then $f(1)=0$, contrdiction If $c=1$ then we get $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ (which indeed fits since $LHS=2xy=RHS$) If $c=-2$ then we get $\boxed{\text{S3 : }f(x)=-2x\quad\forall x}$ (which indeed fits since $LHS=4xy-8xy=-2xy-2xy=RHS$)

The answer is $f(x)\equiv 0,f(x)=x,f(x)=-2x$. If $f(x)\equiv c$, then $x=cx+c,\forall x\in \mathbb{R}$, so $c=0$. If $f$ is non - constant. Let $P(x,y)$ be the assertion in the hypothesis. $P(x;1)$ gives $$f(f(x)+f(x)f(1))=xf(1)+f(x),\forall x\in \mathbb{R}.$$If $f(1)=0$ then $P(1;y)$ gives $f(0)=2f(y)$, so $f$ is constant, this is a contradiction. Then $f(1)\neq 0$, now if there exists $a,b\in \mathbb{R}$ such that $f(a)=f(b)$, from (1), put $x\to a$ and $x\to b$, compare two equations, we have $af(1)=bf(1)$, or $a=b$. Thus, $f$ is injective. $P(1;y)$ gives $f(f(1)\cdot (f(y)+y))=2f(y),\forall y\in \mathbb{R}$. Let $g(x)=f(x)+x$, then $g:\mathbb{R}\to\mathbb{R}$ and $$f(f(1)g(x))=2f(x),\forall x\in \mathbb{R}.$$So if there exists $c,d\in \mathbb{R}$ such that $g(a)=g(b)$, then from (2), put $x\to a$ and $x\to b$, compare two equations, we have $f(a)=f(b)$, then $a=b$ (since $f$ is injective). Thus, $g$ is also an injective function. From problem, change the role of $x$ and $y$, we have $$f(xf(y)+f(x)f(y))=yf(x)+f(xy),\forall x,y\in \mathbb{R}.$$It leads to $$f(yf(x)+f(x)f(y))+yf(x)=f(xf(y)+f(x)f(y))+xf(y),\forall x,y\in \mathbb{R}.$$Or $$g(yf(x)+f(x)f(y))=g(xf(y)+f(x)f(y)),\forall x,y\in \mathbb{R}.$$Since $g$ is injective, then $yf(x)=xf(y),\forall x,y\in \mathbb{R}$, so there exists a constant real number $c\neq 0$ such that $f(x)=cx,\forall x\in \mathbb{R}$. Put $f(x)=cx$ to the problem, we get $$c^2xy+c^3xy=cxy+cxy,\forall x,y\in \mathbb{R},$$then $c^3+c^2=2c$, solve the equation, notice that $c\neq 0$, we get $c\in \left\lbrace 1;-2 \right\rbrace$.