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