Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y$ in $\mathbb{R}$,

$$
(x-y) f(x+y)-(x+y) f(x-y)=4 x y\left(x^{2}-y^{2}\right)
$$
To clarify the problem a bit, we set $x+y=a$ and $x-y=b$. Thus, $x^{2}-y^{2}=a b$, $2 x=a+b$, and $2 y=a-b$. We obtain for all real numbers $a, b$:

$$
b f(a)-a f(b)=a b\left(a^{2}-b^{2}\right)
$$

By taking $a=0$, $b f(0)=0$ hence $f(0)=0$.

Next, for $a, b \neq 0$, we divide by $a b$:

$$
\frac{f(a)}{a}-\frac{f(b)}{b}=a^{2}-b^{2}
$$

which simplifies to

$$
\frac{f(a)}{a}-a^{2}=\frac{f(b)}{b}-b^{2}
$$

This quantity is independent of the chosen parameter! We set $c=\frac{f(a)}{a}-a^{2}$ for a non-zero $a$ (thus for all non-zero $a$)

Thus, for all $x \neq 0$, $f(x)=x^{3}+c x$, but this remains valid for $x=0$ since $f(0)=0$.

All that remains is to verify that these functions work!