20.1. (New York, 78). The function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies the identity

$$
f(x y) \equiv \frac{f(x)+f(y)}{x+y}, \quad x, y \in \mathbf{R}, \quad x+y \neq 0
$$

Does there exist a value $x \in \mathbb{R}$ for which $f(x) \neq 0$?
20.1. Let $y=1$ in the identity satisfied by the function $f(x)$. Then we have the identity

$$
f(x)=(f(x)+f(1)) /(x+1) \quad(x \neq-1)
$$

i.e., $x f(x)=f(1)$. For $x=0$, we get $f(1)=0$, and thus for $x \notin \{-1 ; 0\}$ we have $f(x)=0$. Next, substituting the values $y=0, x=2$ into the original identity, we get $f(0)=(f(2)+f(0)) / 2$, from which $f(0)=$ $=f(2)=0$. Finally, substituting the values $y=0, x=-1$, we get $f(0)=-f(-1)-f(0)$, from which $f(-1)=-2 f(0)=0$. Thus, for the function $f(x)$, the identity $f(x)=0, x \in \mathbf{R}$ has been proven.