$y'=u$
$\large\frac{d^2y}{dx^2}=\frac{du}{dx}$
$\qquad= \large\frac{du}{dy}. \frac{dy}{dx}$
$\qquad= u. \large\frac{du}{dy}$
$u. \large\frac{du}{dy}$$+e^{2y}y^3=0$
$\large\frac{du}{u^2}$$=-e^{2y}dy$
$\large\frac{-1}{u}=\frac{1}{2} $$ e^{2y}+c_1$
$\large\frac{dx}{dy}=\large\frac{1}{2}. e^{2y}+c_1$
$dx=\bigg( \large\frac{1}{2}. e^{2y}+c_1 \bigg)$$dy$
$x= \large\frac{1}{4} .e^{2y}+c_1y+c_2$
Hence d is the correct answer.