Answer
$$\frac{{dy}}{{dx}} = \frac{{2x{y^4} + 2{y^3} - y}}{{x - 6{x^2}{y^3} - 6x{y^2}}}$$
Work Step by Step
$$\eqalign{
& \ln \left( {xy + 1} \right) = 2x{y^3} + 4 \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {\ln \left( {xy + 1} \right)} \right) = \frac{d}{{dx}}\left( {2x{y^3} + 4} \right) \cr
& {\text{sum rule for derivatives}} \cr
& \frac{d}{{dx}}\left( {\ln \left( {xy + 1} \right)} \right) = \frac{d}{{dx}}\left( {2x{y^3}} \right) + \frac{d}{{dx}}\left( 4 \right) \cr
& {\text{use the product rule for }}\frac{d}{{dx}}\left( {2x{y^3}} \right) \cr
& \frac{d}{{dx}}\left( {\ln \left( {xy + 1} \right)} \right) = 2x\frac{d}{{dx}}\left( {{y^3}} \right) + 2{y^3}\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( 4 \right) \cr
& {\text{solve the derivatives }} \cr
& \frac{1}{{xy + 1}}\frac{d}{{dx}}\left( {xy + 1} \right) = 2x\left( {3{y^2}} \right)\frac{{dy}}{{dx}} + 2{y^3}\left( 1 \right) + 0 \cr
& \frac{1}{{xy + 1}}\left( {x\frac{{dy}}{{dx}} + y} \right) = 6x{y^2}\frac{{dy}}{{dx}} + 2{y^3} \cr
& \frac{x}{{xy + 1}}\frac{{dy}}{{dx}} + \frac{y}{{xy + 1}} = 6x{y^2}\frac{{dy}}{{dx}} + 2{y^3} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \frac{x}{{xy + 1}}\frac{{dy}}{{dx}} - 6x{y^2}\frac{{dy}}{{dx}} = 2{y^3} - \frac{y}{{xy + 1}} \cr
& \left( {\frac{x}{{xy + 1}} - 6x{y^2}} \right)\frac{{dy}}{{dx}} = 2{y^3} - \frac{y}{{xy + 1}} \cr
& \left( {\frac{{x - 6x{y^2}\left( {xy + 1} \right)}}{{xy + 1}}} \right)\frac{{dy}}{{dx}} = \left( {\frac{{2{y^3}\left( {xy + 1} \right) - y}}{{xy + 1}}} \right) \cr
& \left( {x - 6x{y^2}\left( {xy + 1} \right)} \right)\frac{{dy}}{{dx}} = \left( {2{y^3}\left( {xy + 1} \right) - y} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{2{y^3}\left( {xy + 1} \right) - y}}{{x - 6x{y^2}\left( {xy + 1} \right)}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x{y^4} + 2{y^3} - y}}{{x - 6{x^2}{y^3} - 6x{y^2}}} \cr} $$
