Answer
$$\frac{{dy}}{{dz}} = \frac{{1 - z}}{{\sqrt {{z^2} - 1} }} + {\sec ^{ - 1}}z $$
Work Step by Step
$$\eqalign{
& y = z{\sec ^{ - 1}}z - \sqrt {{z^2} - 1},\,\,\,\,\,\,\,z > 1 \cr
& {\text{find the derivative of }}y{\text{ with respect to }}z \cr
& \frac{{dy}}{{dz}} = \frac{{d\left( {z{{\sec }^{ - 1}}z - \sqrt {{z^2} - 1} } \right)}}{{dz}} \cr
& \frac{{dy}}{{dz}} = \frac{{d\left( {z{{\sec }^{ - 1}}z} \right)}}{{dz}} - \frac{{d\left( {\sqrt {{z^2} - 1} } \right)}}{{dz}} \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dz}} = z\frac{{d\left( {{{\sec }^{ - 1}}z} \right)}}{{dz}} + {\sec ^{ - 1}}z\frac{{d\left( z \right)}}{{dz}} - \frac{{d\left( {\sqrt {{z^2} - 1} } \right)}}{{dz}} \cr
& {\text{use the formula }}\cr
&\frac{{d\left( {{{\sec }^{ - 1}}u} \right)}}{{dx}} = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}\cr
&{\text{ and the chain rule}} \cr
& \frac{{dy}}{{dz}} = z\left( {\frac{1}{{\left| z \right|\sqrt {{z^2} - 1} }}} \right) + {\sec ^{ - 1}}z\left( 1 \right) - \frac{1}{2}{\left( {{z^2} - 1} \right)^{ - 1/2}}\left( {2z} \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{dz}} = \frac{z}{{\left| z \right|\sqrt {{z^2} - 1} }} + {\sec ^{ - 1}}z - \frac{z}{{\sqrt {{z^2} - 1} }} \cr
& {\text{where}}\,\,\,\,\,\,\,z > 1 \cr
& \frac{{dy}}{{dz}} = \frac{1}{{\sqrt {{z^2} - 1} }} + {\sec ^{ - 1}}z - \frac{z}{{\sqrt {{z^2} - 1} }} \cr
& \frac{{dy}}{{dz}} = \frac{{1 - z}}{{\sqrt {{z^2} - 1} }} + {\sec ^{ - 1}}z \cr} $$
