My question is following.
Find all $f:\mathbb R\to\mathbb R$ which satisfies $\forall x,y\in\mathbb R,\ f(f(x+y)-f(x-y))=y^2f(x)$.
At least, the trivial solution $f=0$ can be answer.
At first, I tried $y=0$, then $f(0)=0$.
And I stuck here, so I tried to get injective answer first. then, when $x=0$, then $f(f(y)-f(-y))=0$, so $f(y)=f(-y)$. This contradicts. So there exists no injective answer.
... And? I don't have any idea since then.
Short outline of a solution.
If $f$ isn't identical $0$ the expression $y^2f(x)$ takes on all positive or all negative values. Evaluating the functional equation at $(x,x)$ and $(x,-x)$ we get $f(f(2x))=f(-f(2x))$ and therefore $f(x)=f(-x)\ \forall x$.
Swapping $x$ and $y$ we get $y^2f(x)=x^2f(y)$. This leads to the non-trivial solutions $f(x)=\pm\frac{x^2}{4}$.