Answer
$${\text{Separable}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + {y^2} = x{y^2} \cr
& {\text{subtract }}{y^2}{\text{ from each side}} \cr
& \frac{{dy}}{{dx}} = x{y^2} - {y^2} \cr
& {\text{factor }}{y^2} \cr
& \frac{{dy}}{{dx}} = {y^2}\left( {x - 1} \right) \cr
& {\text{or}} \cr
& \frac{{dy}}{{dx}} = \frac{{x - 1}}{{{y^{ - 2}}}} \cr
& {\text{The equation can be written in the form }}\frac{{dy}}{{dx}} = \frac{{p\left( x \right)}}{{q\left( y \right)}},{\text{ then}} \cr
& {\text{the given equation is separable}} \cr
& \cr
& \frac{{dy}}{{dx}} + {y^2} = x{y^2} \cr
& {\text{divide by }}{y^2} \cr
& \frac{1}{{{y^2}}}\frac{{dy}}{{dx}} + 1 = x \cr
& {\text{The equation cannot be written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\,\,\,\left( {{\text{linear form}}} \right), \cr
& {\text{ then}}{\text{, the given equation is not linear}} \cr} $$
