Answer
$$ h'(z)=-\frac{3}{2}(z+(z+1)^{1/2})^{-5/2}\left(1+\frac{1}{2}(z+1)^{-1/2}\right) .$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ h(z)=(z+(z+1)^{1/2})^{-3/2}$, the derivative $ h'(z)$, by using the chain rule, is given by
$$ h'(z)=-\frac{3}{2}(z+(z+1)^{1/2})^{-5/2}\left(1+\frac{1}{2}(z+1)^{-1/2}\right) .$$
