VMO

Given $x\ne 0$ and $y=0$, we get $$xf(f(x{)}^2)=f(x{)}^3$$ which implies $$f(f(x{)}^2)=\frac{f(x{)}^3}{x},\forall x\ne 0.$$ Substituting into the original equation, for all $x,y\ne 0$ $$(x-y)\left[\frac{f(x{)}^3}{x}-\frac{f(y{)}^3}{y}\right]=(f(x)-f(y))(f(x{)}^2-f(y{)}^2)(1)$$ 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 $$(f(x)-2013x)(f(x{)}^2-2013^{2}x)=0,\forall x<0.$$ On the other hand, for $x<0$ then $f(x{)}^2>2013^{2}x$, 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 $$(f(x)-2013x)(f(x{)}^2+2013^{2}x)=0,\forall x>0.$$ 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 $$\begin{array}{rl}(f(x)-f(y))(f(x{)}^2-f(y{)}^2)& =2013(x-y)(2013^2{x}^2-2013^2{y}^2)\\ & =2013^3(x-y)(x^2-y^2).\end{array}$$ Thus, $f(x)=2013x$ for all real numbers $x$. $\square$