Saudi Arabian Mathematical Competitions

Denote ($*$) as the given condition.
Let $a = f(0)$. Taking $x = y$ in ($*$), we get
$$f(2x) = a + 4 f(x)$$
for any $x \neq 0$. Now, by replacing $x, y$ by $2x, 2y$ in ($*$) and using the above property (here we choose $x, y$ such that $x + y, x - y, x \neq 0$), we get
$$2x[4 f(x + y) - 4 f(x - y)] = 8y[a + 4 f(x)],$$
or
$$f(x + y) - f(x - y) = \frac{a y}{x} + \frac{4 y}{x} f(x).$$
Combining this with ($*$), we get $a = 0$. It follows that $f(0) = 0$ and $f(2x) = 4 f(x)$ for any $x$.
Replacing $y = -x$ in ($*$), we have
$$f(-2x) = 4 f(x) = f(2x)$$
for any $x \neq 0$. Since $f(0) = 0$, we conclude that $f$ is an even function. Now, replacing $x$ by $x$ in (1), we get
$$x[f(-x - y) - f(y - x)] = 4y f(-x)$$
or
$$x[f(y + x) - f(y - x)] = 4y f(x).$$
It follows that
$$x y [f(y + x) - f(y - x)] = 4 y^2 f(x).$$
From ($*$), we get $y [f(y + x) - f(y - x)] = 4x f(y)$. Combine with the above equation, we get
$$x^2 f(y) = y^2 f(x)$$
for any real numbers $x, y$. This shows that
$$f(x) = a x^2$$
where $a$ is a constant, which is truly a solution.