9. (10) Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying

$$
f\left(x^{3}+y^{3}\right)=x^{2} f(x)+y f\left(y^{2}\right)
$$

for all $x, y \in \mathbb{R}$.
Solution: substitute $x=0$ and $y=0 \Rightarrow f(0)=0$

$f\left(x^{3}+0^{3}\right)=x^{2} f(x)$

$f\left(0^{3}+x^{3}\right)=x f\left(x^{2}\right)$

Therefore, $f\left(x^{3}\right)=x^{2} f(x)=x f\left(x^{2}\right)$ and $f\left(x^{3}+y^{3}\right)=f\left(x^{3}\right)+f\left(y^{3}\right)$

From this it follows that $f(x+y)=f(x)+f(y)$

$y=-x \Rightarrow f(-x)=-f(x)$, that is, $f$ is odd. We will further assume that the argument is greater than zero.

Then $x^{2} f(x)=x f\left(x^{2}\right) \Rightarrow f\left(x^{2}\right)=x f(x)$

We get

$f\left((x+1)^{2}\right)=(x+1) f(x+1)=(x+1)(f(x)+f(1))=x f(x)+f(x)+x f(1)+f(1)$

But on the other hand

$f\left((x+1)^{2}\right)=f\left(x^{2}+x+x+1\right)=f\left(x^{2}\right)+f(x)+f(x)+f(1)=x f(x)+2 f(x)+f(1)$

Equating these expressions we get

$x f(x)+f(x)+x f(1)+f(1)=x f(x)+2 f(x)+f(1)$

$f(x)=x f(1)$

That is, $f(x)=a x$. It is clear that all such functions satisfy the condition of the problem.