demmy wrote: Defined all $f : \mathbb{R} \to \mathbb{R} $ that satisfied equation $$P(x, y): f(x)f(y)f(x-y)=x^2f(y)-y^2f(x)$$for all $x,y \in \mathbb{R}$ $P(0, 0) \implies f(0)=0$. Case 1: There exists $t \neq 0$ s.t. $f(t)=0$ Then $P(x,t)$ $\implies$ $t^2f(x)=0$ for all $x$. So $\boxed{f=0}$ which satisfies the FE. Case 2: $f(t)=0 \iff t=0$ Let $x \neq 0$. $P(2x,x) \land P(x, 2x) \implies f(-x)=-f(x)$. $P(2x,x) \land P(x, -x)$ $\implies$ $f(x)^2=x^2$ for all $x \ne 0$. Since $f(0)=0$ we have $f(x)^2=x^2$ for all $x$. If $f$ is a solution, then $-f$ is also a solution, WLOG $f(1)=1$. Suppose there exists $x \neq 0$ s.t. $f(x)=-x$, then $P(x,1) \implies f(x-1)=-x-1$, so $(x+1)^2=(x-1)^2$ which gives $x = 0$ and it contradicts our assumption $x \neq 0$. Therefore $f(x)=x$ for all $x \neq 0$, combined with $f(0)=0$ we get $\boxed{f=\mathrm{Id}}$ which is indeed a solution. We conclude that the solution are $\boxed{f=0}$, $\boxed{f=\mathrm{Id}}$ and $\boxed{f=-\mathrm{Id}}$

Let $P(x,y):=f(x)f(y)f(x-y)=x^2f(y)-y^2f(x)$ $P(0,0)$ yields $f(0)^3=0\Longrightarrow f(0)=0$ Now assume that $\exists\alpha\text{ such that }f(\alpha)=0\text{ and }\alpha\neq0$ Thus, notice that from $P(x,\alpha)$ we obtain $f(x)f(\alpha)f(x-\alpha)=x^2f(\alpha)-\alpha^2f(x)\Longrightarrow-\alpha^2f(x)=0$ which forces $f\equiv0$ which is a solution. So from now on assume that $f\not\equiv0$ $P(x,-x)$ yields $f(x)f(-x)f(2x)=x^2f(-x)-x^2f(x)\Longrightarrow f(x)f(-x)f(2x)=x^2(f(-x)-f(x))$ (1) $P(-x,x)$ yields $f(x)f(-x)f(-2x)=x^2f(x)-x^2f(-x)\Longrightarrow -f(x)f(-x)f(-2x)=x^2(f(-x)-f(x))$ (2) Furthermore from (1) and (2) we obtain $f(x)f(-x)f(2x)=-f(x)f(-x)f(-2x)\Longrightarrow f(2x)=-f(-2x), \forall x\neq0$ Moreover let $x\to\frac{x}{2}$ and we obtain $f(x)=-f(-x)\Longrightarrow-f(x)=f(-x), \forall x\neq0$ however since $f(0)=0$, we get $-f(x)=f(-x), \forall x\in\mathbb{R}$ and $f$ is odd. Now from $P(2x,x)$ we obtain $f(2x)f(x)^2=4x^2f(x)-x^2f(2x)$ (!) and from $P(x,2x)$ we obtain $f(x)f(-x)f(2x)=x^2f(2x)-4x^2f(x)$ (?) Furthermore, using the fact that $f$ is odd and examining $P(2x,x)+P(x,-x)$ we obtain $$f(2x)f(x)^2+f(x)f(-x)f(2x)=4x^2f(x)-x^2f(x)-x^2f(2x)+x^2f(-x)-x^2f(x)\Longrightarrow f(2x)f(x)^2-f(x)^2f(2x)=3x^2f(x)-x^2f(2x)+x^2f(-x)\Longleftrightarrow3x^2f(x)-x^2f(2x)-x^2f(x)=0\Longrightarrow2x^2f(x)=x^2f(x)$$Thus $2f(x)=f(2x)$, however since $f(0)=0$ we have $2f(x)=f(2x), \forall x\in\mathbb{R}$ Now plugging our last result into (?) we obtain $$-f(x)^2f(2x)=2x^2f(x)-4x^2f(x)\Longrightarrow-f(x)^2f(2x)=-2x^2f(x)\Longrightarrow-2f(x)^3=-2x^2f(x)\Longrightarrow f(x)^3=x^2f(x)\Longrightarrow f(x)^2=x^2, \forall x\neq0$$which forces $f(x)=x\text{ and }f(x)=-x, \forall x\neq0$ however since $f(0)=0$ we obtain $f(x)=x\text{ and }f(x)=-x, \forall x\in\mathbb{R}$ So, to sum up $\boxed{f(x)=x, f(x)=-x\text{ and }f\equiv0, \forall x\in\mathbb{R}}$ $\blacksquare$.

Sketch of my solution: Firstly note that $\boxed{f\equiv0 \forall x\in\mathbb{R}}$ is a solution, then assume $f\not\equiv0$. $P(0,0)$ yields $f(0)=0$ and we can easily get that $f$ is injective at $0$. $P(1,-1), P(1,2), P(2,1)$ implies that $f(1)^2=1, f(2)=2f(1), f(1)+f(-1)=0$ Case1) $f(1)=1, f(-1)=-1, f(2)=2.$ Note that for the substitutions $x \neq 0,1,-1,2$ Comparing $P(x,1)$ and $P(x,x-1)$ implies that $f(x)^2(x-2)+2xf(x)-x^3=0 \implies f(x)=x$ or $f(x)=-\frac{x^2}{x-2}$ If there exists $t$ such $f(t)=-\frac{t^2}{t-2}$, then $P(t,2)$ yields contradiction. So $\boxed{f(x)=x \forall x\in\mathbb{R}}$ Case2) $f(1)=-1, f(-1)=1, f(2)=-2.$ Similarly, we get $\boxed{f(x)=-x \forall x\in\mathbb{R}} \ \ \ \ \ \ \ \ \ \blacksquare$