Answer
$$\frac{{dy}}{{dx}} = \frac{{y{e^y}\cos x}}{{1 - y{e^y}\sin x}}$$
Work Step by Step
$$\eqalign{
& \ln y = {e^y}\sin x \cr
& {\text{find the derivative using the implicit differentiation;}} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {{e^y}\sin x} \right] \cr
& {\text{use product rule}} \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = {e^y}\frac{d}{{dx}}\left[ {\sin x} \right] + \sin x\frac{d}{{dx}}\left[ {{e^y}} \right] \cr
& {\text{solving the derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = {e^y}\left( {\cos x} \right) + \sin x\left( {{e^y}} \right)\frac{{dy}}{{dx}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = {e^y}\cos x + {e^y}\sin x\frac{{dy}}{{dx}} \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} - {e^y}\sin x\frac{{dy}}{{dx}} = {e^y}\cos x \cr
& \frac{{dy}}{{dx}} - y{e^y}\sin x\frac{{dy}}{{dx}} = y{e^y}\cos x \cr
& \left( {1 - y{e^y}\sin x} \right)\frac{{dy}}{{dx}} = y{e^y}\cos x \cr
& \frac{{dy}}{{dx}} = \frac{{y{e^y}\cos x}}{{1 - y{e^y}\sin x}} \cr} $$
