Answer
$$g'\left( z \right) = \frac{{3{z^2}\left( {1 - {z^2}} \right)}}{{{{\left( {1 + {z^2}} \right)}^4}}}$$
Work Step by Step
$$\eqalign{
& g\left( z \right) = {\left( {\frac{z}{{1 + {z^2}}}} \right)^3} \cr
& {\text{Differentiate}} \cr
& g'\left( z \right) = \frac{d}{{dz}}\left[ {{{\left( {\frac{z}{{1 + {z^2}}}} \right)}^3}} \right] \cr
& {\text{Use the chain rule and the power rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr
& g'\left( z \right) = 3{\left( {\frac{z}{{1 + {z^2}}}} \right)^3}^{ - 1}\frac{d}{{dz}}\left[ {\frac{z}{{1 + {z^2}}}} \right] \cr
& {\text{By the quotient rule}} \cr
& g'\left( z \right) = 3{\left( {\frac{z}{{1 + {z^2}}}} \right)^2}\left( {\frac{{\left( {1 + {z^2}} \right)\left( 1 \right) - z\left( {2z} \right)}}{{{{\left( {1 + {z^2}} \right)}^2}}}} \right) \cr
& {\text{Multiply and simplify}} \cr
& g'\left( z \right) = 3{\left( {\frac{z}{{1 + {z^2}}}} \right)^2}\left( {\frac{{1 + {z^2} - 2{z^2}}}{{{{\left( {1 + {z^2}} \right)}^2}}}} \right) \cr
& g'\left( z \right) = 3{\left( {\frac{z}{{1 + {z^2}}}} \right)^2}\left( {\frac{{1 - {z^2}}}{{{{\left( {1 + {z^2}} \right)}^2}}}} \right) \cr
& g'\left( z \right) = \frac{{3{z^2}\left( {1 - {z^2}} \right)}}{{{{\left( {1 + {z^2}} \right)}^4}}} \cr} $$
