Answer
$$ y'
=- \frac{1}{(z-1)^2}\sqrt{\frac{z-1}{z+1}}.$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\sqrt{\frac{z+1}{z-1}}$, by using the chain and product rules, the derivative $ y'$ is given by
$$ y'=\frac{1}{2\sqrt{\frac{z+1}{z-1}}}\left( \frac{(z-1)(1)-(z+1)(1)}{(z-1)^2}\right)\\
=- \frac{1}{(z-1)^2}\sqrt{\frac{z-1}{z+1}}.$$
