Answer
$$f'\left( y \right) = \frac{{4y}}{{1 + {{\left( {2{y^2} - 4} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( y \right) = {\tan ^{ - 1}}\left( {2{y^2} - 4} \right) \cr
& {\text{find the derivative}} \cr
& f'\left( y \right) = \frac{d}{{dy}}\left[ {{{\tan }^{ - 1}}\left( {2{y^2} - 4} \right)} \right] \cr
& {\text{use }}\frac{d}{{dy}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dy}}.{\text{ let }}u = 2{y^2} - 4 \cr
& f'\left( y \right) = \frac{1}{{1 + {{\left( {2{y^2} - 4} \right)}^2}}}\frac{d}{{dy}}\left[ {2{y^2} - 4} \right] \cr
& f'\left( y \right) = \frac{1}{{1 + {{\left( {2{y^2} - 4} \right)}^2}}}\left( {4y} \right) \cr
& {\text{simplifying}} \cr
& f'\left( y \right) = \frac{{4y}}{{1 + {{\left( {2{y^2} - 4} \right)}^2}}} \cr} $$
