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

$$
\forall x, y \in \mathbb{R}, \quad f(x+y)+x y=f(x) f(y)
$$
By setting $y=0$, we obtain for all $x \in \mathbb{R}: f(x)=f(x) \cdot f(0)$. Since the zero function is not a solution, there exists $x_{0} \in \mathbb{R}$ such that $f\left(x_{0}\right) \neq 0$. Since $f\left(x_{0}\right)=f\left(x_{0}\right) \cdot f(0)$, we deduce that $f(0)=1$.

Furthermore, for $y=-x$, we have $1-x^{2}=f(x) f(-x)$, so in particular for $x=1$, we have $0=$ $f(1) f(-1)$. We then have two cases to consider. If $f(1)=0$, then by taking $y=1, f(x+1)+x=0$, so for all $x \in \mathbb{R}, f(x)=1-x$. If $f(-1)=0$, then for $y=-1$ we have $f(x-1)-x=0$, so for all $x \in \mathbb{R}, f(x)=x+1$. Conversely, we verify that the functions $x \rightarrow x+1$ and $x \rightarrow 1-x$ are indeed solutions, and these are the only ones.