Selection tests for the Gulf Mathematical Olympiad 2013

Let $f:\mathbb{R}\to \mathbb{R}$ be a function which satisfies the two functional equations. Because $f\left(\frac{\sqrt{3}}{3}x\right)-\sqrt{3}f(1)=-\frac{2\sqrt{3}}{3}\ne 0$, either $f(1)\ne 0$ or $f\left(\frac{\sqrt{3}}{3}x\right)\ne 0$. Fix $x_0\in \mathbb{R}$ with $f\left(x_0\right)\ne 0$ and let $y\in \mathbb{R}$ with $y\ne 0$. We have $$f\left(x_0\right)f\left(\frac{1}{y}\right)=f\left(\frac{x_0}{y}\right)+f\left(x_{0}y\right)=f\left(x_0\right)f(y).$$ Therefore, $f\left(\frac{1}{y}\right)=f(y)$ for all $y\ne 0$. Let $x\ne 0$. We have $$\begin{array}{rl}f\left(\frac{\sqrt{3}}{3}x\right)& =\sqrt{3}f(x)-\frac{2\sqrt{3}}{3}x\\ & =\sqrt{3}f\left(\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}x}\right)-\frac{2\sqrt{3}}{3}x\\ & =\sqrt{3}\left(\sqrt{3}f\left(\frac{1}{\frac{\sqrt{3}}{3}x}\right)-\frac{2}{x}\right)-\frac{2\sqrt{3}}{3}x\\ & =3f\left(\frac{\sqrt{3}}{3}x\right)-\frac{2\sqrt{3}}{x}-\frac{2\sqrt{3}}{3}x\end{array}$$ We deduce that $$f\left(\frac{\sqrt{3}}{3}x\right)=\frac{\sqrt{3}}{x}+\frac{\sqrt{3}}{3}x$$ and therefore $$f(x)=x+\frac{1}{x}$$ for all $x\ne 0$. For $x=0$ and $y=2$, we have from the second functional equation $$\frac{5}{2}f(0)=f(0)+f(0)$$ which implies that $f(0)=0$. Hence $$f(x)=\{\begin{array}{cr}x+\frac{1}{x}& \text{ if }x\ne 0\\ 0& \text{ otherwise }\end{array}$$ Conversely, it is easy to check that this function satisfies the two functional equations.