Answer
$$\frac{{dy}}{{dx}} = \frac{{2{e^{2x}} - \cos \left( {x + 3y} \right)}}{{3\cos \left( {x + 3y} \right)}}$$
Work Step by Step
$$\eqalign{
& {e^{2x}} = \sin \left( {x + 3y} \right) \cr
& {\text{find the derivative using implicit differentiation; }} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {{e^{2x}}} \right] = \frac{d}{{dx}}\left[ {\sin \left( {x + 3y} \right)} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{ and }}\frac{d}{{dx}}\left[ {\sin u} \right] = \cos u\frac{{du}}{{dx}} \cr
& {e^{2x}}\frac{d}{{dx}}\left[ {2x} \right] = \cos \left( {x + 3y} \right)\frac{d}{{dx}}\left[ {x + 3y} \right] \cr
& {\text{solving the derivatives}} \cr
& {e^{2x}}\left( 2 \right) = \cos \left( {x + 3y} \right)\left( {1 + 3\frac{{dy}}{{dx}}} \right) \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& 2{e^{2x}} = \cos \left( {x + 3y} \right) + 3\cos \left( {x + 3y} \right)\frac{{dy}}{{dx}} \cr
& 3\cos \left( {x + 3y} \right)\frac{{dy}}{{dx}} = 2{e^{2x}} - \cos \left( {x + 3y} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{2{e^{2x}} - \cos \left( {x + 3y} \right)}}{{3\cos \left( {x + 3y} \right)}} \cr} $$
