Macedonian Mathematical Competitions

If $x = y$, we have
$$f(x^2 - y^2) = (x - y)[f(x) + f(y)], \quad f(0) = 0 \cdot [2f(x)] = 0.$$
Then $f(0) = 0$.
If $y = -x$, we have
$$\begin{aligned} f[x^2 - (-x)^2] &= [x - (-x)][f(x) + f(-x)], \quad f(0) = 2x[f(x) + f(-x)], \\ 0 &= -2x[f(x) + f(-x)]. \end{aligned}$$
From the last equation we have $f(x) + f(-x) = 0$, $f(-x) = -f(x)$, for every $x \in \mathbb{R}$. This means that $f$ is an odd function.

$(x - y)[f(x) + f(y)] = f(x^2 - y^2) = (x + y)[f(x) - f(y)]$,
i.e.
$$(x-y)[f(x)+f(y)]=(x+y)[f(x)-f(y)], \\ xf(x)-yf(x)+xf(y)-yf(y)=xf(x)+yf(x)-xf(y)-yf(y), \\ -yf(x)+xf(y)=yf(x)-xf(y), \\ 2yf(x)=2xf(y).$$
If $y=1$ and $f(1)=k$, then $f(x)=kx$, for $k \in \mathbb{R}$. It's easy to see that all the functions $f(x)=kx$ are the solution to (1).