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