Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

$$
f(x) f(y) + f(x+y) = x y
$$
By setting $y=0$, we have $f(x)(f(0)-1)=0$. Thus, either $f(0)=-1$, or $f$ is identically zero. The second case is impossible, for example, because we would have $0=1$ by substituting $x=y=1$. Therefore, $f(0)=-1$. To use this information, we substitute $y=-x=-1$ into the initial equation, which becomes $f(1) f(-1)=0$. We will handle the two cases separately:

- Case $1: f(1)=0$. We want to nullify the product $f(x) f(y)$, so we set $y=1$, and we see that we need to replace $x$ with $x-1$ to get $f(x)=x-1$.
- Case $2: f(-1)=0$. Similarly, by setting $y=-1$ and replacing $x$ with $x+1$, we have $f(x)=-x-1$.

We verify that these two affine functions are solutions.