Answer
$f'\left( z \right) = \frac{{2z{e^{\arcsin \left( {{z^2}} \right)}}}}{{\sqrt {1 - {z^4}} }}$
Work Step by Step
$$\eqalign{
& f\left( z \right) = {e^{\arcsin \left( {{z^2}} \right)}} \cr
& {\text{Differentiate}}{\text{, apply }}\frac{d}{{dz}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dz}} \cr
& f'\left( z \right) = \frac{d}{{dz}}\left[ {{e^{\arcsin \left( {{z^2}} \right)}}} \right] \cr
& f'\left( z \right) = {e^{\arcsin \left( {{z^2}} \right)}}\frac{d}{{dz}}\left[ {\arcsin \left( {{z^2}} \right)} \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }} \cr
& f'\left( z \right) = {e^{\arcsin \left( {{z^2}} \right)}}\left( {\frac{1}{{\sqrt {1 - \left( {{z^2}} \right)} }}} \right)\frac{d}{{dz}}\left[ {{z^2}} \right] \cr
& f'\left( z \right) = {e^{\arcsin \left( {{z^2}} \right)}}\left( {\frac{1}{{\sqrt {1 - \left( {{z^2}} \right)} }}} \right)\left( {2z} \right) \cr
& {\text{Simplify}} \cr
& f'\left( z \right) = \frac{{2z{e^{\arcsin \left( {{z^2}} \right)}}}}{{\sqrt {1 - {z^4}} }} \cr} $$
