1. Find all functions $f: \mathbb{R} \backslash\{-1\} \rightarrow \mathbb{R}$, for which

$$
f(x)+f(y)=(x+y+2) f(x) f(y)
$$

for all $x, y \in \mathbb{R} \backslash\{-1\}$.
IV/1. We substitute $x=y=0$ into the functional equation to get $f(0)=f(0)^{2}$. Therefore, $f(0)=0$ or $f(0)=1$. If $f(0)=0$, we substitute $y=0$ into the functional equation and get $f(x)=0$ for every $x$. Clearly, this function satisfies the conditions of the problem. If $f(0)=1$, we again substitute $y=0$ into the functional equation and derive $f(x)=\frac{1}{x+1}$. We can verify that this function also satisfies the conditions of the problem. The solutions to the problem are therefore the functions $f(x)=0$ and $f(x)=\frac{1}{x+1}$.
Writing the equality for $x=0, y=0$ or $x=y$ ..... 1 point
Determining that $f(0)=0$ or $f(0)=1$ ..... 2 points
Determining that $f(x)=0$ is a solution ..... 1 point
Determining that $f(x)=\frac{1}{x+1}$ is a solution ..... 1 point
Determining that there are no other solutions ..... 1 point
Verifying that $f(x)=\frac{1}{x+1}$ is indeed a solution (it is obvious that $f(x)=0$ is a solution) ..... 1 point