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

$$
f(x) f(y) + f(x+y) = xy
$$
By choosing $y=0$ we have

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

and thus $f(0)=-1$ or $f(x)=0$ for all $x$. The zero function is not a solution to the exercise, which would give $0=x y$ for all $x, y \in \mathbb{R}$. Therefore, $f(0)=-1$. With $x=1$ and $y=-1$ we have

$$
f(1) f(-1)-1=-1 \Rightarrow f(1) f(-1)=0
$$

Thus, $f(1)=0$ or $f(-1)=0$. If $f(1)=0$, with $y=1$, we have

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

so $f(x)=x-1$. If $f(-1)=0$, with $y=-1$ we have

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

so $f(x)=-x-1$. We verify that these two solutions are possible

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

and

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

In conclusion, the two possible solutions are indeed $f(x)= \pm x-1$