13. Find $f: \mathbf{R} \rightarrow \mathbf{R}$, for any $x, y \in \mathbf{R}$, such that $f\left(x^{2}-y^{2}\right)=(x-y)(f(x)+f(y))$.
13. Let $x=y$, then $f(0)=0$. Let $x=-1, y=0$, then $f(1)=-f(-1)$. Let $x=a, y=1$ and $x=a, y=-1$. Then we have $f\left(a^{2}-1\right)=(a-1)(f(a)+f(1)), f\left(a^{2}-1\right)=(a+1)(f(a)-f(1))$. Therefore, $f(a)=a f(1)$. Here $a \in \mathbf{R}$. Thus, any function satisfying the given conditions is $f(x)=k x$, where $k$ is a constant. Upon verification, it satisfies the original functional equation.