Answer
$${\text{The equation is separable}}$$
Work Step by Step
$$\eqalign{
& {t^2}y'\left( t \right) = \left( {t + 4} \right){y^2} \cr
& {\text{Multiply both sides of the equation by }}\frac{1}{{{t^2}{y^2}}} \cr
& \frac{1}{{{t^2}{y^2}}}\left[ {{t^2}y'\left( t \right)} \right] = \frac{1}{{{t^2}{y^2}}}\left[ {\left( {t + 4} \right){y^2}} \right] \cr
& \frac{1}{{{y^2}}}y'\left( t \right) = \frac{1}{{{t^2}}}\left( {t + 4} \right){y^2} \cr
& {\text{The d}}{\text{.e can be written in the form }}g\left( y \right)y'\left( t \right) = h\left( t \right),{\text{ then}} \cr
& {\text{we can conclude that the equation is separable}}{\text{.}} \cr} $$