Answer
$$\frac{{dy}}{{dx}} = \frac{{3{x^2} - 2x{e^y}}}{{{x^2}{e^y} + 1}}$$
Work Step by Step
$$\eqalign{
& {x^2}{e^y} + y = {x^3} \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{x^2}{e^y} + y} \right) = \frac{d}{{dx}}\left( {{x^3}} \right) \cr
& {\text{use sum rule as follows}} \cr
& \frac{d}{{dx}}\left( {{x^2}{e^y}} \right) + \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {{x^3}} \right) \cr
& {\text{use the product rule}} \cr
& {x^2}\frac{d}{{dx}}\left( {{e^y}} \right) + {e^y}\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {{x^3}} \right) \cr
& {\text{solve the derivatives using the chain rule and the power rule}} \cr
& {x^2}{e^y}\frac{{dy}}{{dx}} + {e^y}\left( {2x} \right) + \frac{{dy}}{{dx}} = 3{x^2} \cr
& {x^2}{e^y}\frac{{dy}}{{dx}} + 2x{e^y} + \frac{{dy}}{{dx}} = 3{x^2} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& {x^2}{e^y}\frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} = 3{x^2} - 2x{e^y} \cr
& \left( {{x^2}{e^y} + 1} \right)\frac{{dy}}{{dx}} = 3{x^2} - 2x{e^y} \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^2} - 2x{e^y}}}{{{x^2}{e^y} + 1}} \cr} $$
