Answer
$g'\left( z \right) = \frac{{\sec z + \tan z - z\sec z\tan z - z{{\sec }^2}z}}{{{{\left( {\sec z + \tan z} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& g\left( z \right) = \frac{z}{{\sec z + \tan z}} \cr
& {\text{Differentiating}} \cr
& g'\left( z \right) = \frac{d}{{dz}}\left[ {\frac{z}{{\sec z + \tan z}}} \right] \cr
& {\text{Use the quotient rule for derivatives }}\left( {\frac{f}{h}} \right)' = \frac{{hf' - fh'}}{{{h^2}}} \cr
& {\text{Let }}f = z{\text{ and }}h = \sec z + \tan z,{\text{ then}} \cr
& g'\left( z \right) = \frac{{\left( {\sec z + \tan z} \right)\left( z \right)' - z\left( {\sec z + \tan z} \right)'}}{{{{\left( {\sec z + \tan z} \right)}^2}}} \cr
& {\text{Use the Derivatives of Trigonometric Functions }} \cr
& g'\left( z \right) = \frac{{\left( {\sec z + \tan z} \right)\left( 1 \right) - z\left( {\sec z\tan z + {{\sec }^2}z} \right)}}{{{{\left( {\sec z + \tan z} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& g'\left( z \right) = \frac{{\sec z + \tan z - z\sec z\tan z - z{{\sec }^2}z}}{{{{\left( {\sec z + \tan z} \right)}^2}}} \cr}
$$
