Find all functions $f: \mathbb{R}_{+}^{*} \rightarrow \mathbb{R}_{+}^{*}$ such that $f\left(x^{2}+x f(y)\right)=x f(x+y)$ for all real numbers $x, y>0$.
The idea here, as in many functional equations, is to make sure that expressions being the argument of $f$ are equal, so that the images are equal.

Here, we want $x^{2}+x f(y)=x+y$ with $x, y>0$, which, given the conditions, would ensure $x=1$, and then, thanks to $x^{2}+x f(y)=x+y$, we get $f(y)=y$.

So, let's fix a $y>0$; if we find $x>0$ such that $x^{2}+x f(y)=x+y$, we have $f(y)=y$ as previously stated. It therefore suffices to show the existence of such an $x$. The quadratic polynomial $X^{2}+(f(y)-1)X-y$ is 0, and $f(y)=y$; the only possible solution is therefore the identity. It is clear that this is a solution conversely.