Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$:

$$
f(x) f(y)=f(x-y)
$$
Considering $x=y=0$, we have $f(0)^{2}=f(0)$, so $f(0)=0$ or 1.

Assume $f(0)=0$, consider $y=0$, we have $f(x)=0$, thus, $f=0$, which is valid.

Otherwise, $f(0)=1$, take $x=y$, we have $f(x)^{2}=1$, hence $f(x)= \pm 1$. Note that this does not mean the function is constantly 1 or -1. Consider $x=0$, we have $f(x)=f(-x)$, so $f$ is even. Next, take $y=-x$, to get

$$
1=f(x)^{2}=f(x) f(-x)=f(2 x)
$$

For a real number $y$, we can consider $x=\frac{y}{2}$, and we have $f(y)=1$, thus $f=1$ which is also valid.