Answer
$$y = \pm \sqrt {2{t^3} + C} $$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{{3{t^2}}}{y} \cr
& {\text{separate}} \cr
& ydy = 3{t^2}dt \cr
& {\text{integrate both sides}} \cr
& \int {ydy} = \int {3{t^2}} dt \cr
& \frac{{{y^2}}}{2} = {t^3} + c \cr
& {y^2} = 2{t^3} + 2c \cr
& {\text{solve for }}y \cr
& y = \pm \sqrt {2{t^3} + 2c} \cr
& {\text{set }}C = 2c \cr
& y = \pm \sqrt {2{t^3} + C} \cr} $$