Thailand Mathematical Olympiad

First, we show that $f(ab)=f(a)+f(b)$ for $a,b\in (0,\infty )$ such that $a^{2}b\ge 4$. In fact, if $a^{2}b\ge 4$ and $a,b>0$, then we can find $x,y>0$ such that $x+y=ab$ and $xy=b$, namely $$x=\frac{a+\sqrt{a^2-(4/b)}}{(2/b)}>0\text{and}y=\frac{a-\sqrt{a^2-(4/b)}}{(2/b)}>0.$$

Thus, for $a,b\in (0,\infty )$ such that $a^{2}b\ge 4$, we plug $x,y$ as above in the functional equation to get $f(ab)=f(x+y)=f(\frac{x+y}{xy})+f(xy)=f((ab)/b)+f(b)=f(a)+f(b)$. Now let $x,y>0$. Let $$z=\max \left\{\frac{4}{x^2{y}^2},\frac{4}{x^{2}y},\frac{4}{y^2}\right\}>0.$$ Then $(xy{)}^{2}z\ge 4$, $x^2(yz)\ge 4$ and $y^{2}z\ge 4$ which imply that $f(xyz)=f(xy)+f(z)$, $f(xyz)=f(x)+f(yz)$ and $f(yz)=f(y)+f(z)$ respectively. Therefore, $$ $$\begin{array}{rl}f(xy)& =f(xyz)-f(z)\\ & =(f(x)+f(yz))-f(z)\\ & =f(x)+f(yz)-f(z)\\ & =f(x)+(f(y)+f(z))-f(z)=f(x)+f(y).\end{array}$$