Thai Mathematical Olympiad

Define $\lambda (x,y)=\frac{2008}{x^2{y}^3}$ for all $x,y>0$. We can see that ${\left(\frac{x}{\lambda (x,y)}\right)}^2(y\cdot \lambda (x,y){)}^3=2008$. Thus, $${\left(\frac{x}{z}\right)}^2(yz{)}^3>2008\text{ for every }z>\lambda (x,y).$$ It follows that $(f(xy){)}^2=f(x^2/z^2)f(y^2{z}^2)$ for all $z>\lambda (x,y)$. Now for any positive real numbers $x$ and $y$, we choose $$z>\max \{\lambda (x,x),\lambda (y,y),\lambda (x,y),\lambda (y,x)\}.$$ It follows that $$(f(x^2){)}^2=f(x^2/z^2)f(x^2{z}^2)(\because z>\lambda (x,x))$$ $$(f(y^2){)}^2=f(y^2/z^2)f(y^2{z}^2)(\because z>\lambda (y,y))$$ $$(f(xy){)}^2=f(x^2/z^2)f(y^2{z}^2)(\because z>\lambda (x,y))$$ $$(f(xy){)}^2=f(y^2/z^2)f(x^2{z}^2)(\because z>\lambda (y,x))$$ We can see that $$\begin{array}{rl}f(xy{)}^2\cdot f(xy{)}^2& =(f(x^2/z^2)f(y^2{z}^2))(f(y^2/z^2)f(x^2{z}^2))\\ & =(f(x^2/z^2)f(x^2{z}^2))(f(y^2/z^2)f(y^2{z}^2))\\ & =(f(x^2){)}^2(f(y^2){)}^2\end{array}$$ Thus $(f(xy){)}^2=f(x^2)f(y^2)$ for all $x,y>0$.