hi

question reads

find in implicit form the general solution of differential equation

$\displaystyle \frac{dy}{dx}=3y^2e^{-2x}\sqrt{8+e^{-2x}}$

=$\displaystyle \int{3y^2}=\int{e^{-2x}}\sqrt{8+e^{-2x}}$

=$\displaystyle \int{3y^2}=\int{e^{-2x}}+e^{\frac{-3}{2}}+8^{\frac{1}{2}}$

$\displaystyle y^3=\frac{-1}{2}e^{-2x}-\frac{2}{3}e^{\frac{3}{2}}+\frac{2}{3}8x^\frac{3}{ 2}$

i did try but as you can see not sure what i am doing could do with some expert help please