Vietnamese Mathematical Olympiad

Given $x \neq 0$ and $y = 0$, we get
$$x f(f(x)^2) = f(x)^3$$
which implies
$$f(f(x)^2) = \frac{f(x)^3}{x}, \forall x \neq 0.$$
Substituting into the original equation, for all $x, y \neq 0$
$$(x - y) \left[ \frac{f(x)^3}{x} - \frac{f(y)^3}{y} \right] = (f(x) - f(y))(f(x)^2 - f(y)^2) \quad (1)$$
Substituting $x < 0, y = 1$ into (1), we have
$$(x - 1) \left[ \frac{f(x)^3}{x} - 2013^3 \right] = (f(x) - 2013)(f(x)^2 - 2013^2),$$
which is equivalent to
$$(f(x) - 2013x)(f(x)^2 - 2013^2x) = 0, \forall x < 0.$$
On the other hand, for $x < 0$ then $f(x)^2 > 2013^2x$, it follows that $f(x) = 2013x$ for all $x < 0$. Hence $f(-1) = -2013$.

Replacing $x > 0$, $y = -1$ into (1), we get
$$(x+1) \left[ \frac{f(x)^3}{x} - 2013^3 \right] = (f(x)+2013)(f(x)^2 - 2013^2),$$
which is equivalent to
$$(f(x) - 2013x)(f(x)^2 + 2013^2x) = 0, \forall x > 0.$$
or
$$f(x) = 2013x, \forall x > 0.$$
Combining with $f(0) = 0$, we get $f(x) = 2013x$ for all real numbers $x$. With $f(x) = 2013x$, we get $f(x)^2 = 2013^2 x^2$ and
$$f(f(x)^2) = f((2013x)^2) = 2013^3 x^2.$$
Hence, we can get with the given conditions
$$(x-y)(f(f(x)^2)-f(f(y)^2)) = 2013^3(x-y)(x^2-y^2)$$
and
$$\begin{aligned} (f(x)-f(y))(f(x)^2-f(y)^2) &= 2013(x-y)(2013^2x^2-2013^2y^2) \\ &= 2013^3(x-y)(x^2-y^2). \end{aligned}$$
Thus, $f(x) = 2013x$ for all real numbers $x$. $\square$