Note that $y''' = 2y'e^{2y}$ which is precisely $y''' = 2y'y''$

The RHS is just $\dfrac{d}{dx}(y')^2$ therefore you can integrate both sides and obtain that

$y'' = (y')^2 + 1$ where the constant is due to initial conditions.

Hence, use the original ODE again to get $(y')^2 = e^{2y} - 1$

Hence $y' = \pm \sqrt{e^y-1}$.

Now just separate variables and integrate to obtain

$\arctan(\sqrt{e^{2y} - 1}) = x $ where $|x| < \dfrac{\pi}{2}$

Hence $e^{2y}-1 = \tan^2 x$ and so $y = \dfrac{1}{2}\log \sec^2 x$

i.e. $y = \log |\sec x|$