The zero constant function is a solution, but we exclude this case. By setting $x=0$, we get $f(0)=0$, and by setting $y=-1$, we get $f(-1)=-1$. By setting $x=-1$ and $y=-\frac{1}{2}$, we get $f\left(-\frac{1}{2}\right)=-\frac{1}{2}$. Finally, by setting $x=-\frac{1}{2}$ and $y=1$, we get $f(1)=1$. Then, by setting $x=1$, we obtain $f(y+1)=1+f(y)$, so $f(x(y+1))=f(x) f(y+1)$. By making a change of variable, we have that $f$ is multiplicative. The original equation then gives:

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

By varying $y$, we get that $f$ is additive. Moreover, since $f$ is multiplicative, for $y \geqslant 0$, we have $f(y)=f(\sqrt{y})^{2} \geqslant 0$. Therefore, $f(x+y)=f(x)+f(y) \geqslant f(x)$, so $f$ is increasing and additive with $f(1)=1$, hence it is the identity (according to the study of Cauchy's equation).
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying:

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