Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy:

$$
\forall x, y \in \mathbb{R}, \quad f(x f(y)+y)=f(f(x y))+y
$$
By setting $x=0$, we obtain for all $y \in \mathbb{R}: f(y)=f(f(0))+y$. Thus, for all $y \in \mathbb{R}$, we have $f(y)=y+C$, with $C \in \mathbb{R}$, and the initial equation becomes: for all $x, y \in \mathbb{R}, x y+x C+y+C=$ $x y+2 C+y$, so $C=0$ (by setting $x=0$ for example). Conversely, the identity function is indeed a solution.