Answer
$$y = - \frac{1}{{\left( {1/2} \right){e^{2x}} + C}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {y^2}{e^{2x}} \cr
& {\text{Separating variables leads to}} \cr
& \frac{{dy}}{{{y^2}}} = {e^{2x}} \cr
& {\text{use }}\frac{1}{{{a^n}}} = {a^{ - n}} \cr
& {y^{ - 2}}dy = {e^{2x}} \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {{y^{ - 2}}dy} = \int {{e^{2x}}dx} \cr
& {\text{rewrite the integrand on the right side of the equation}} \cr
& \int {{y^{ - 2}}dy} = \frac{1}{2}\int {{e^{2x}}\left( 2 \right)dx} \cr
& {\text{integrating by using }}\int {{e^u}du} = {e^u} + C{\text{ and the power rule }} \cr
& \frac{{{y^{ - 1}}}}{{ - 1}} = \frac{1}{2}{e^{2x}} + C \cr
& {\text{solve the equation for }}y \cr
& - \frac{1}{y} = \frac{1}{2}{e^{2x}} + C \cr
& y = - \frac{1}{{\left( {1/2} \right){e^{2x}} + C}} \cr} $$
