Ans: $f(x)=x+\frac{1}{x}.$ It is easy to see that the above function is a solution. From $f(x)f(y)=f(xy)+f\left(\frac{x}{y}\right)$, let $y\rightarrow \frac{1}{y}$, the RHS is fixed but the LHS gives $f(y)=f\left(\frac{1}{y}\right)$ for all real $y$ since $f$ is not equal to $0$ for all real numbers. Now, let $P(x)$ be the the assertion $f \left(\frac{\sqrt3}{3} x\right) = \sqrt3 f(x) - \frac{2\sqrt3}{3} x$, \[P\left(\frac{1}{x}\right)\implies f(\sqrt 3 x)=f\left(\frac{\sqrt{3}}{3x}\right)=\sqrt{3}f\left(\frac{1}{x}\right)-\frac{2\sqrt 3}{3x}=\sqrt{3}f(x)-\frac{2\sqrt 3}{3x}\]but we know $f(\sqrt 3 x)=f(x)f(\sqrt{3})-f\left(\frac{x}{\sqrt 3}\right)=f(x)f(\sqrt{3})-f\left(\frac{\sqrt 3}{x}\right)$. Thus, we now have a system of simultaneous equations relating to $f(x)$ and $f\left(\frac{\sqrt 3}{x}\right)$, \[f\left(\frac{\sqrt 3}{x}\right)=f \left(\frac{\sqrt3}{3} x\right) = \sqrt3 f(x) - \frac{2\sqrt3}{3} x\]\[-f\left(\frac{\sqrt 3}{x}\right)=(\sqrt 3-f(\sqrt 3))f(x)-\frac{2\sqrt 3}{3x}\]and adding these two equations together yields $f(x)=C\left(x+\frac{1}{x}\right)$ where $C\in \mathbb{R}$. Plugging the function into $P(x)$ will yield $f(x)=x+\frac{1}{x}$ as the only solution. $\blacksquare$