Determine all functions from $\mathbb{R}$ to $\mathbb{R}$ satisfying:

$$
\forall(x, y) \in \mathbb{R}^{2}, f(x) f(y) + f(x+y) = x y
$$
Let $y=0$. We have $f(x)(f(0)+1)=0$. We proceed by case analysis:

- If $f(0) \neq -1$, then $\forall x \in \mathbb{R}, f(x)=0$.
- If $f(0)=-1$. Then, with $y=-x, f(x) f(-x)=1-x^{2}$. So with $x=1$, we have:

$-f(1)=0$. Therefore, with $y=1, f(x+1)=x$ or $f(x)=x-1$.

$-f(-1)=0$. Therefore, with $y=-1, f(x-1)=-x$ or $f(x)=-x-1$.

In the end, it is easily verified that $f: x \mapsto 0, f: x \mapsto -x-1, f: x \mapsto x-1$ are indeed solutions.