Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that

\[
f\left(x^{2} y\right)=f(x y)+y f(f(x)+y)
\]

for all real numbers \( x \) and \( y \).
Answer: $f(y)=0$.

By substituting $x=0$ into the original equation we obtain $f(0)=f(0)+y f(f(0)+y)$, which after simplifying yields $y f(f(0)+$ $y)=0$. This has to hold for every $y$.

Now let's substitute $y=-f(0)$. We get that $-(f(0))^{2}=0$, which gives us $f(0)=0$.

By plugging the last result into $y f(f(0)+y)=0$ we now get that $y f(y)=0$.

Therefore if $y \neq 0$ then $f(y)=0$.

Altogether we have shown that $f(y)=0$ for every $y$ is the only possible solution, and it clearly is a solution.