Answer
$$\frac{{dy}}{{dx}} = \frac{{6\cos x}}{{{{\left( {3 - 2\sin x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{2\sin x}}{{3 - 2\sin x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\frac{{2\sin x}}{{3 - 2\sin x}}} \right] \cr
& {\text{use the quotient rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {3 - 2\sin x} \right) \cdot {D_x}\left( {2\sin x} \right) - 2\sin x \cdot {D_x}\left( {3 - 2\sin x} \right)}}{{{{\left( {3 - 2\sin x} \right)}^2}}} \cr
& {\text{solve derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {3 - 2\sin x} \right)\left( {2\cos x} \right) - 2\sin x\left( { - 2\cos x} \right)}}{{{{\left( {3 - 2\sin x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{6\cos x - 4\sin x\cos x + 4\sin x\cos x}}{{{{\left( {3 - 2\sin x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{6\cos x}}{{{{\left( {3 - 2\sin x} \right)}^2}}} \cr} $$
