Find the minimum value of $f$ 
Find the minimum value of $$f(x)=\frac{\tan \left(x+\frac{\pi}{6}\right)}{\tan x}, \qquad x\in \left(0,\frac{\pi}{3}\right).$$

My approach is as follows. I tried to solve it by segregating it  
$$f(x)=\frac{1}{\sqrt{3}\tan x}+\left({\sqrt{3}}+\frac{1}{\sqrt{3}}\right)\frac{1}{\sqrt{3}-\tan x},$$ but $f'(x)$ is getting more and more complicated.

We know that $\tan(x)\tan(y)=1$ then $x+y=\frac{\pi}{2}$ thus we have $f(x)=\frac{1}{\tan(x)\tan(\frac{\pi}{3}-x)}$ thus we need to maximize denominator (lll be referring to it as k)to get minimum value . Now using the fact that $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ we have $\tan(\frac{\pi}{3})=\frac{\tan(\frac{\pi}{3}-x)+\tan(x)}{1-k}$ thus $k=g(x)=1-\frac{1}{\sqrt{3}}(tan(\frac{\pi}{3}-x)+\tan(x))$ thus maximum value is at $g'(x)=\sec^2(\frac{\pi}{3}-x)-\sec^2(x)=0$ in the given domain we have only one solution at $x=\frac{\pi}{6}$ thus the minimum value of $f(x)=3$
