Given $f: k \rightarrow R$, for all $x, y \in \mathbf{R}$, it satisfies
$$
f\left(x^{2}-y^{2}\right)=x f(x)-y f(y) .
$$

Find $f(x)$.
Let $y=0$ we get
$$
f\left(x^{2}\right)=x f(x)
$$

Let $x=0$ we get
$$
f\left(-y^{2}\right)=-y f(y)=-f\left(y^{2}\right)
$$

Therefore, $f(x)$ is an odd function, so we only need to consider the case $x \geqslant 0$.
Let $a=x^{2}, b=y^{2}$, then by the given and (1)
$$
f(a-b)=f(a)-f(b)
$$

Using the usual Cauchy method, we know that (2) implies
$$
f(x)=k x, x \in \mathbf{Q}
$$

where $k=f(1)$.
For any positive real number $x$, take a rational number $y$, then by (1) and (3)
$$
f\left((x-y)^{2}\right)=(x-y) f(x-y)=(x-y)(f(x)-k y)
$$

On the other hand
$$
f\left((x-y)^{2}\right)=f\left(x^{2}-2 x y+y^{2}\right)=f\left(x^{2}\right)-2 f(x y)+f\left(y^{2}\right)
$$

By (2), it is easy to see that for $y \in \mathbf{Q}, f(x y)=y f(x)$, so
$$
f\left((x-y)^{2}\right)=x f(x)-2 y f(x)+k y^{2} .
$$

Let $y \rightarrow x$, comparing the two expressions of $f\left((x-y)^{2}\right)$ above, we get
$$
f(x)=k x .
$$

Therefore, (4) always holds.
Clearly, the $f(x)$ in (4) satisfies the conditions.