Let $f$ be a real valued function, defined on  $R-\left\{-1,1\right\}$ and given by  $f\left(x\right)=3{\log }_e\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}.$  Then in which of the following intervals, function $f\left(x\right)$  is increasing?

$$\begin{gathered} \begin{array}{c}\text{Given }f\left(x\right)=3{\log }_e\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}.\\ \text{Differentiate both sides with respect to }x\\ f^{\text{'}}\left(x\right)=3\frac{x-1}{x+1}\times \frac{x+1-x+1}{{\left(x+1\right)}^2}+\frac{2}{{\left(x+1\right)}^2}\\ =\frac{2}{x+1}\left(\frac{3}{x+1}+\frac{1}{x-1}\right)\\ =\frac{4}{{\left(x-1\right)}^2\left(x+1\right)}\left(2x-1\right)\end{array} \\ The function f\left(x\right) is increa\sin g when \left(2x-1\right)\left(x+1\right)>0 and x not equal to 1 \\ x\in \left(-\infty ,-1\right)\cup \left[\frac{1}{2},\infty \right)-\left\{1\right\} \end{gathered}$$