The Differential Equation: $\displaystyle 3y^2 y''(x) + 2e^y y'(x)+2y y'(x)+6y(y'(x))^2=0$ is a perfect derivative. Find the first integral.

$\displaystyle 3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0$

not sure which notation is best so i put 2 ways of the same problem.

This is the product rule split up but i just cannot see how to do this backwards. Any help would be great! Thanks.