2. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ that satisfy $x f(y)+y f(x)=(x+y) \cdot f(x) \cdot f(y), x, y \in \mathbf{R}$.
(1968 Bulgarian Competition Problem)
2. In the given condition, let $x=y=1$, then we get $2 f(1)=2[f(1)]^{2}$, i.e., $f(1)=0$ or $f(1)=1$. If $f(1)=0$, then in the given condition, let $y=1$, we get $f(x)=0$. If $f(1)=1$, then let $y=-1$, we get $f(x)+x=(x+1) \cdot f(x)$, thus, they satisfy the conditions of the problem.