Answer
$\frac{dy}{dz}={\frac{1-z}{\sqrt{(z^2-1)}}}+sec^{-1}{z}$
Work Step by Step
Given $y=z\sec^{-1}{z}-\sqrt{(z^2-1)}$
On differentiating both sides:
$\frac{dy}{dz}=\frac{d(z\sec^{-1}{z}-\sqrt{(z^2-1)})}{dz}$
$\frac{dy}{dz}={\frac{z}{|z|\sqrt{(z^2-1)}}}+sec^{-1}{z}-\frac{1}{2\sqrt{(z^2-1)}}\frac{d({(z^2-1)})}{dz}$
$\frac{dy}{dz}={\frac{z}{|z|\sqrt{(z^2-1)}}}+sec^{-1}{z}-\frac{z}{\sqrt{(z^2-1)}}$
$\frac{dy}{dz}={\frac{1-z}{\sqrt{(z^2-1)}}}+sec^{-1}{z}$