Answer
$$y'=\frac{-2}{(x+1)(x-1)}$$
Work Step by Step
$$y=\log_{3}\Big(\Big(\frac{x+1}{x-1}\Big)^{\ln3}\Big)$$
We have $\log_ax^b=b\log_ax$. Therefore, $$y=\ln3\log_3\Big(\frac{x+1}{x-1}\Big)$$
The derivative of $y$ is $$y'=\ln3\Big(\log_3\Big(\frac{x+1}{x-1}\Big)\Big)'$$
Recall Theorem 7: $$\frac{d}{dr}\log_au=\frac{1}{u\ln a}\frac{du}{dr}$$
That means: $$y'=\ln3\times\frac{1}{\frac{x+1}{x-1}\ln3}\Big(\frac{x+1}{x-1}\Big)'=\frac{1}{\frac{x+1}{x-1}}\times\frac{(x-1)-(x+1)}{(x-1)^2}$$
$$y'=\frac{x-1}{x+1}\times\frac{x-1-x-1}{(x-1)^2}=\frac{x-1}{x+1}\times\frac{-2}{(x-1)^2}$$
$$y'=\frac{-2}{(x+1)(x-1)}$$
