Answer
$$y'\left( x \right) = \frac{{y\cos x}}{{{e^y} - 1 - \sin x}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{e^y}}}{{1 + \sin x}} \cr
& {\text{use implicit differentiation differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{{e^y}}}{{1 + \sin x}}} \right) \cr
& {\text{use quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + \sin x} \right)\frac{d}{{dx}}\left( {{e^y}} \right) - {e^y}\frac{d}{{dx}}\left[ {1 + \sin x} \right]}}{{{{\left( {1 + \sin x} \right)}^2}}} \cr
& {\text{solving derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + \sin x} \right)\left( {{e^y}} \right)\frac{{dy}}{{dx}} - {e^y}\left( {0 + \cos x} \right)}}{{{{\left( {1 + \sin x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + \sin x} \right)\left( {{e^y}} \right)\frac{{dy}}{{dx}} - {e^y}\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}} \cr
& {\left( {1 + \sin x} \right)^2}\frac{{dy}}{{dx}} = \left( {1 + \sin x} \right)\left( {{e^y}} \right)\frac{{dy}}{{dx}} - {e^y}\cos x \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& {\left( {1 + \sin x} \right)^2}\frac{{dy}}{{dx}} - \left( {1 + \sin x} \right)\left( {{e^y}} \right)\frac{{dy}}{{dx}} = - {e^y}\cos x \cr
& \left( {{{\left( {1 + \sin x} \right)}^2} - \left( {1 + \sin x} \right)\left( {{e^y}} \right)} \right)\frac{{dy}}{{dx}} = - {e^y}\cos x \cr
& \frac{{dy}}{{dx}} = \frac{{ - {e^y}\cos x}}{{{{\left( {1 + \sin x} \right)}^2} - \left( {1 + \sin x} \right)\left( {{e^y}} \right)}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^y}\cos x}}{{\left( {1 + \sin x} \right)\left( {{e^y}} \right) - {{\left( {1 + \sin x} \right)}^2}}} \cr
& {\text{or we can write}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^y}\cos x}}{{\left( {{e^y} - 1 - \sin x} \right)\left( {1 + \sin x} \right)}} \cr
& y'\left( x \right) = \frac{{{e^y}}}{{1 + \sin x}}\left( {\frac{{\cos x}}{{{e^y} - 1 - \sin x}}} \right) \cr
& y'\left( x \right) = y\left( {\frac{{\cos x}}{{{e^y} - 1 - \sin x}}} \right) \cr
& y'\left( x \right) = \frac{{y\cos x}}{{{e^y} - 1 - \sin x}} \cr} $$
