3. Find all real functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f\left(x^{2}-y^{2}\right)=(x-y)[f(x)+f(y)]
$$
Solution. Let $f$ be a function that is a solution to equation (1). If we choose $x=y$, we get

$$
\begin{aligned}
& f\left(x^{2}-x^{2}\right)=(x-x)[f(x)+f(x)] \\
& f(0)=0 \cdot[2 f(x)]=0
\end{aligned}
$$

Thus, $f(0)=0$.

If in equation (1) we replace $y$ with $-x$, we get

$$
\begin{aligned}
& f\left[x^{2}-(-x)^{2}\right]=[x-(-x)][f(x)+f(-x)] \\
& f(0)=2 x[f(x)+f(-x)] \\
& 0=-2 x[f(x)+f(-x)]
\end{aligned}
$$

Since the last equality holds for every $x \in \mathbb{R}$, we get

$$
\begin{aligned}
& f(x)+f(-x)=0 \\
& f(-x)=-f(x)
\end{aligned}
$$

Thus, $f$ is an odd function.

If we replace $y$ with $-y$, and from the fact that $f$ is an odd function, equation (1) becomes

$$
\begin{aligned}
& f\left(x^{2}-(-y)^{2}\right)=[x-(-y)][f(x)+f(-y)] \\
& f\left(x^{2}-y^{2}\right)=(x+y)[f(x)-f(y)]
\end{aligned}
$$

From (1) and (2) we get

$$
(x-y)[f(x)+f(y)]=f\left(x^{2}-y^{2}\right)=(x+y)[f(x)-f(y)]
$$

or

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

$$
\begin{aligned}
& -y f(x)+x f(y)=y f(x)-x f(y) \\
& 2 y f(x)=2 x f(y)
\end{aligned}
$$

If we choose $y=1$ and introduce the notation $f(1)=k$, we get $f(x)=k x$, where $k \in \mathbb{R}$. It is not difficult to verify that every function of the form $f(x)=k x$ is a solution to equation (1).