If $f$ is a solution then $-f$ is also a solution so we may assume $f(1)\geq 0$(Wrong position) Let $P(x,y): f(xf(x+y))+f((x+y)f(y))=(x+y)^2$ and $Q(x,y)=P(x-y,y): f((x-y)f(x))+f(xf(y))=x^2$ $P(0,0): f(0)=0$ $P(x,0): f(xf(x))=x^2: R(x)$ If $f(x_0)=0$ then $x_0^2=f(x_0f(x_0))=f(0)=0$ so $x_0=0$ If there are $0\neq x_1\neq x_2\neq 0$ such that $f(x_1)=f(x_2)$ then $Q(x_1,x_2): f((x_1-x_2)f(x_1))+f(x_1f(x_2))=x_1^2\Leftrightarrow $ $\Leftrightarrow f((x_1-x_2)f(x_1))+f(x_1f(x_1))=f(x_1f(x_1))\Leftrightarrow (x_1-x_2)f(x_1)=0\Leftrightarrow x_1=0$ a contradiction. So $f$ is $1-1$ $R(-x):f(-xf(x))=(-x)^2=x^2=f(xf(x))\Leftrightarrow f(-x)=-f(x)$ If $f$ is a solution then $-f$ is also a solution so we may assume $f(1)\geq 0$(Correct position) $R(1): f(f(1))=1$ and $R(f(1)): f(f(1)f(f(1)))=f(1)^2\Leftrightarrow f(f(1))=f(1)^2$ and since $f(1)>0$ we have that $f(1)=1$ $Q(x,-x): f(2xf(x))-f(xf(x))=x^2\Leftrightarrow f(2xf(x))=2x^2=(\sqrt{2}x)^2=f(\sqrt{2}xf(\sqrt{2}x))\Leftrightarrow$ $\Leftrightarrow f(\sqrt{2}x\sqrt{2})=\sqrt{2}f(x\sqrt{2})=2f(x)\Leftrightarrow f(2x)=2f(x)$ $Q(1,x): -f((x-1)f(1))+f(f(x))=1\Leftrightarrow f(f(x))=1+f(x-1)$ $Q(1,2x): 1+f(2x-1)=f(f(2x))=f(2f(x))=2f(f(x))=2+2f(x-1)=2+f(2x-2)$ $2x\rightarrow x$: $ 1+f(x-1)=2+f(x-2)$ so $ f(x+1)=f(x)+1$ so $f(f(x))=1+f(x-1)=f(x)$ and since $f$ is $1-1$ we have that $f(x)=x\forall x$ So we have that the solutions are $f(x)=x,\forall x\in \mathbb{R}$ and $f(x)=-x,\forall x\in \mathbb{R}$

Prod55 wrote: If $f$ is a solution then $-f$ is also a solution so ... I dont see why ... Thanks for any explanation.

pco wrote: Prod55 wrote: If $f$ is a solution then $-f$ is also a solution so ... I dont see why ... Thanks for any explanation. Since $f(-x)=-f(x)$ we have that $-f(x(-f(x+y))-f((x+y)(-f(y))=f(xf(x+y))+f((x+y)f(y))=(x+y)^2$ so $-f$ is also a solution. I had to mention first that $f(-x)=-f(x)$, i am apologies for this , but I did not use the fact that $f(1)\geq 0$ to prove that $f(-x)=-f(x)$ so i think that my proof is complete.

Let $P(x,y)$ denote $f(x(f(x+y)))+f((x+y)f(y))=(x+y)^2$ $P(0,0)$ -> $f(0)=0$ $P(x,0)$ -> $f(xf(x))=x^2$ If there is a real number $c$ such that $f(c)=0$, $P(c,0)$ -> $f(cf(c))=c^2$ -> $c^2=0$ -> $c=0$ So, $f(c)=0$ if and only if $c=0$. $P(x+y,-x)$ -> $f((x+y)f(y))+f(yf(-x))=y^2$ $P(-y,x+y)$ -> $f(-yf(x))+f(xf(x+y))=x^2$ Sum these two equations up, we get $[f(y(f(-x)))+f(-yf(x))]+[f(x(f(x+y)))+f((x+y)f(y))]=x^2+y^2$ So, $f(y(f(-x)))+f(-yf(x))=-2xy$ Plug $x=1$ -> $f(yf(-1))+f(-yf(1))=-2y$ (*) $P(1,0)$ -> $f(f(1))=1$ $P(f(1),0)$ -> $f(f(1)f(f(1)))=f(1)^2$ -> $f(1)=1$ or $f(1)=-1$ Case 1: $f(1)=1$ If there is a real number $c$ such that $f(c)=1$, $P(c,0)$ -> $f(cf(c))=c^2$ -> $c=1$ or $c=-1$ $P(-1,0)$ -> $f(-f(-1))=1$ -> $-f(-1)=1$ or $-f(-1)=-1$ So, $f(-1)=1$ or $f(-1)=-1$ If $f(-1)=1$, $P(2,-1)$ -> $f(2f(1))+f(f(-1))=1$ -> $f(2)=0$ which contradicts to $f(c)=0$ if and only if $c=0$. Hence, $f(-1)=-1$ From (*), $f(-y)=-y$ -> $f(x)=x$ for all real number $x$. Case 2: $f(1)=-1$ $P(1,0)$ -> $f(f(1))=1$ -> $f(-1)=1$ From (*), $f(y)=-y$ -> $f(x)=-x$ for all real number $x$. It's clear that these two solutions satisfy our assertion. Hence, $f(x)=x$ and $f(x)=-x$ are the only two solutions.

Very similar with IMO 2009 SL A7

Let $P(x, y)$ be the given assertion. $P(0, 0)\implies\boxed{f(0)=0}$ and $P(0, x)\implies\boxed{f(xf(x))=x^2}$. Claim: $f$ is injective. Proof: If $f(x)=f(y)$, $P(x-y, y)\implies f((x-y)f(x))+f(x\underbrace{f(y)}_{f(x)})=x^2\Rightarrow f((x-y)f(x))=0$. But $f(t)=0\Rightarrow t^2=f(tf(t))=0\Rightarrow t=0$. Thus $x=y$ or $f(x)=0\rightarrow f(y)=0\rightarrow x=y=0$, so $x=y \ $ $_{\blacksquare}$ That means that $xf(x)=(-x)f(-x)\rightarrow\underbrace{\boxed{f(x)=-f(-x)}}_{(*)}$. $$P(x, y)\implies \textcolor{red}{f(xf(x+y))}+\textcolor{blue}{f((x+y)f(y))}=(x+y)^2$$$$P(-y, x+y)\implies \textcolor{green}{f(-yf(x))}+\textcolor{red}{f(xf(x+y))}=x^2$$$$P(-x-y, x)\implies \underbrace{\textcolor{blue}{f(-(x+y)f(-y))}}_{(*)}+\textcolor{green}{f(-yf(x))}=y^2$$ Solving the system yields $f(xf(x+y))=x(x+y)$. If $f(1)=c$, we know that $f(c)=1$, so plugging $y=c-x$, we get $\boxed{f(x)=x\cdot c}$. But plugging it again, $P(x, y)$ gives us $c^2=1\implies \boxed{f(x)=x}$ or $\boxed{f(x)=-x} \ $ $_{\blacksquare}$