Answer
$-\dfrac{w}{(w^2-1)^{3/2}}$
Work Step by Step
Consider $z(w) = \dfrac{1}{\sqrt {(w^{2}-1)}}$
We will use the derivative definition, such as
$z'(w) = \lim\limits_{h \to 0}\dfrac{\dfrac{1}{\sqrt {((w+h)^{2}}-1)}-\dfrac{1}{\sqrt {(w^{2}}-1)}}{h}= \lim\limits_{h \to 0}\dfrac{-(h+2w)}{(\sqrt {((w+h)^{2}-1)} \sqrt{(w^{2}-1)})(\sqrt {((w+h)^{2}-1)} +\sqrt {(w^{2}-1))}}$
and $\dfrac{-(0+2w)}{(\sqrt {((w+0)^{2}-1)} \sqrt{(w^{2}-1)})(\sqrt {((w+0)^{2}-1)} +\sqrt {(w^{2}-1))}}=\dfrac{-w}{(w^2-1)\sqrt{w^2-1}}=-\dfrac{w}{(w^2-1)^{3/2}}$