Answer
$$ y'=\frac{ 1 -\frac{z}{2}}{(1-z)^{3/2}}.$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(x^n)'=nx^{n-1}$
Since $ y= \frac{z}{\sqrt{1-z}}$, then by the quotient rule the derivative $ y'$ is given by
$$ y'=\frac{\sqrt{1-z} +\frac{z}{2\sqrt{1-z}}}{1-z} =\frac{ 1-z +\frac{z}{2}}{(1-z)\sqrt{1-z}}=\frac{ 1 -\frac{z}{2}}{(1-z)^{3/2}}.$$
