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