Answer
$$\frac{{dy}}{{dx}} = \frac{{2x{y^4} - y}}{{1 - 3{x^2}{y^3}}}$$
Work Step by Step
$$\eqalign{
& x + \ln y = {x^2}{y^3} \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {x + \ln y} \right) = \frac{d}{{dx}}\left( {{x^2}{y^3}} \right) \cr
& {\text{use sum rule as follows}} \cr
& \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {\ln y} \right) = \frac{d}{{dx}}\left( {{x^2}{y^3}} \right) \cr
& {\text{use the product rule}} \cr
& \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {\ln y} \right) = {x^2}\frac{d}{{dx}}\left( {{y^3}} \right) + {y^3}\frac{d}{{dx}}\left( {{x^2}} \right) \cr
& {\text{solve the derivatives}} \cr
& 1 + \frac{1}{y}\frac{{dy}}{{dx}} = {x^2}\left( {3{y^2}} \right)\frac{{dy}}{{dx}} + {y^3}\left( {2x} \right) \cr
& 1 + \frac{1}{y}\frac{{dy}}{{dx}} = 3{x^2}{y^2}\frac{{dy}}{{dx}} + 2x{y^3} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} - 3{x^2}{y^2}\frac{{dy}}{{dx}} = 2x{y^3} - 1 \cr
& \left( {\frac{1}{y} - 3{x^2}{y^2}} \right)\frac{{dy}}{{dx}} = 2x{y^3} - 1 \cr
& {\text{multiply both sides by }}y \cr
& \left( {1 - 3{x^2}{y^3}} \right)\frac{{dy}}{{dx}} = 2x{y^4} - y \cr
& \frac{{dy}}{{dx}} = \frac{{2x{y^4} - y}}{{1 - 3{x^2}{y^3}}} \cr} $$
