Answer
$\frac{{dy}}{{dx}} = \frac{{{y^3}}}{{y + 1 - 2x{y^2}}}$
Work Step by Step
$$\eqalign{
& y + \ln y = x{y^2} \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \frac{d}{{dx}}\left[ y \right] + \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {x{y^2}} \right] \cr
& {\text{Use the product rule on the right hand side}} \cr
& \frac{d}{{dx}}\left[ y \right] + \frac{d}{{dx}}\left[ {\ln y} \right] = x\frac{d}{{dx}}\left[ {{y^2}} \right] + {y^2}\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Compute the derivatives}}{\text{, using the chain rule for differentiation}} \cr
& \frac{{dy}}{{dx}} + \frac{1}{y}\frac{{dy}}{{dx}} = x\left( {2y} \right)\frac{{dy}}{{dx}} + {y^2}\left( 1 \right) \cr
& \frac{{dy}}{{dx}} + \frac{1}{y}\frac{{dy}}{{dx}} = 2xy\frac{{dy}}{{dx}} + {y^2} \cr
& {\text{Collect the }}\frac{{dy}}{{dx}}{\text{ terms}} \cr
& \frac{{dy}}{{dx}} + \frac{1}{y}\frac{{dy}}{{dx}} - 2xy\frac{{dy}}{{dx}} = {y^2} \cr
& {\text{Factor and solve for }}\frac{{dy}}{{dx}} \cr
& \left( {1 + \frac{1}{y} - 2xy} \right)\frac{{dy}}{{dx}} = {y^2} \cr
& \frac{{dy}}{{dx}} = \frac{{{y^2}}}{{1 + \frac{1}{y} - 2xy}} \cr
& \frac{{dy}}{{dx}} = \frac{{{y^3}}}{{y + 1 - 2x{y^2}}} \cr} $$
