Determine the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f\left(x^{2}-y^{2}\right)=(x-y)(f(x)+f(y))
$$
By setting $x=0, y=0$, we have $f(0)=0$. For $y=-x$, it follows that $f(0)=2 x(f(x)+f(-x))$ from which $\mathrm{f}(\mathrm{x})$ $=0$ for all $x \neq 0$. By substituting $-y$ for $y$, the equation becomes

$$
\begin{gathered}
f\left(x^{2}-(-y)^{2}\right)=(x+y)(f(x)+f(-y)) \\
f\left(x^{2}-y^{2}\right)=(x+y)(f(x)-f(y))
\end{gathered}
$$

Combining this equation with the initial functional equation:

$$
\begin{aligned}
(x-y)(f(x)+f(y)) & =(x+y)(f(x)-f(y)) \\
x f(y) & =y f(x)
\end{aligned}
$$

and by taking $y=1$, we get $f(x)=x f(1)$. We are thus looking for solutions of the form $f(x)=a x:$

$$
a\left(x^{2}-y^{2}\right)=(x-y)(a x+a y)
$$

so $f(x)=a x$ is a solution for all $a$.