, 2013

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function which satisfies the two functional equations.
Because $f\left(\frac{\sqrt{3}}{3} x\right)-\sqrt{3} f(1)=-\frac{2 \sqrt{3}}{3} \neq 0$, either $f(1) \neq 0$ or $f\left(\frac{\sqrt{3}}{3} x\right) \neq 0$.
Fix $x_{0} \in \mathbb{R}$ with $f\left(x_{0}\right) \neq 0$ and let $y \in \mathbb{R}$ with $y \neq 0$. 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 $y \neq 0$.
Let $x \neq 0$. We have
$$\begin{aligned} f\left(\frac{\sqrt{3}}{3} x\right) & =\sqrt{3} f(x)-\frac{2 \sqrt{3}}{3} x \\ & =\sqrt{3} f\left(\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3} x}\right)-\frac{2 \sqrt{3}}{3} x \\ & =\sqrt{3}\left(\sqrt{3} f\left(\frac{1}{\frac{\sqrt{3}}{3} x}\right)-\frac{2}{x}\right)-\frac{2 \sqrt{3}}{3} x \\ & =3 f\left(\frac{\sqrt{3}}{3} x\right)-\frac{2 \sqrt{3}}{x}-\frac{2 \sqrt{3}}{3} x \end{aligned}$$
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 $x \neq 0$.
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
$$f(x)=\left\{\begin{array}{cr} x+\frac{1}{x} & \text{ if } x \neq 0 \\ 0 & \text{ otherwise } \end{array}\right.$$
Conversely, it is easy to check that this function satisfies the two functional equations.