Answer
$$\frac{{dy}}{{dx}} = \frac{{4x}}{{1 + {{\left( {2{x^2} - 3} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \arctan \left( {2{x^2} - 3} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\arctan \left( {2{x^2} - 3} \right)} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {\arctan u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {2{x^2} - 3} \right)}^2}}}\frac{d}{{dx}}\left[ {2{x^2} - 3} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {2{x^2} - 3} \right)}^2}}}\left( {4x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{4x}}{{1 + {{\left( {2{x^2} - 3} \right)}^2}}} \cr} $$
