Answer
$$\frac{d y}{d x}= \frac{-2}{(x+1)(x-1)}\\
$$
Work Step by Step
Given $$ y=\log _{3}\left(\left(\frac{x+1}{x-1}\right)^{\ln 3}\right)$$
Since $$\log_{a}z=\frac{\ln z}{\ln a}$$ So, we have
\begin{aligned}
y&=\frac{\ln \left(\frac{x+1}{x-1}\right)^{\ln 3}}{\ln 3}\\
&=\frac{(\ln 3) \ln\left(\frac{x+1}{x-1}\right)}{\ln 3}\\
&=\ln \left(\frac{x+1}{x-1}\right)\\
&=\ln (x+1)-\ln (x-1)\\
&
\Rightarrow \frac{d y}{d x}=\frac{1}{x+1}-\frac{1}{x-1}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{x-1-x-1}{(x+1)(x-1)}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{-2}{(x+1)(x-1)}\\
\end{aligned}
