34.5. Find all functions $f(x)$ that are defined for all $x$ and satisfy the relation

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

for all $x, y$.
34.5. Let $x=y$. Then we get $2 x f(x)=2 x(f(x))^{2}$. If $x \neq 0$, then $f(x)=0$ or 1.

Suppose that $f(a)=0$ for some $a \neq 0$. Setting $x=a$, we get $a f(y)=0$ for all $y$, i.e., $f=0$.

Suppose that $f(a)=1$ for some $a \neq 0$. Setting $x=a$, we get $a f(y)+y=(a+y) f(y)$, i.e., $y=y f(y)$. Thus, $f(y)=1$ for all $y \neq 0$.

As a result, we obtain that either $f(x)=0$ for all $x$, or $f(x)=1$ for all $x \neq 0$ and $f(0)=c-$ an arbitrary number.