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

$$
f(x) f(y) + f(x+y) = xy
$$

for all real numbers $x, y$.
By replacing $y$ with 0 in the functional equation, we obtain

$$
f(x)(f(0)+1)=0
$$

If $f(0) \neq -1$, then for all real $x$, $f(x)=0$. However, conversely, if for all real $r$, $f(r)=0$, then by setting $x=y=1$ for example, the equation becomes $0=1$, which is absurd. Thus, we have that $f(0)=-1$. By setting $y=1$ and $x=-1$ in the equation, we have:

$$
f(-1) f(1) + f(0) = -1
$$

which simplifies to: $f(-1) f(1) = 0$. We then have two cases, depending on whether $f(1)=0$ or $f(-1)=0$.

Case 1: $f(1)=0$. We then set $y=1$ in the functional equation:

$$
f(1) f(x) + f(x+1) = x
$$

By replacing $x$ with $x-1$ and using that $f(1)=0$, we have $f(x) = x-1$ for all real $x$. Conversely, the function that maps $x$ to $x-1$ is indeed a solution to the equation since for all real $x, y$:

$$
(x-1)(y-1) + (x+y-1) = xy
$$

Case 2: $f(-1)=0$. We proceed as in the previous case: we set $y=-1$ to obtain:

$$
f(-1) f(x) + f(x-1) = -x
$$

Then for all $x$, $f(x-1) = -x$ and by replacing $x-1$ with $x$, we have for all $x$: $f(x) = -x-1$. Conversely, the function that maps $x$ to $-x-1$ is indeed a solution.

The solutions are therefore: $f(x) = x-1$ for all real $x$, and: $f(x) = -x-1$ for all real $x$.