First Order ODE/y - x y' = y' y^2 exp y

Theorem
The first order ODE:
 * $(1): \quad y - x y' = y' y^2 e^y$

has the solution:
 * $x y^2 = e^y + C$

Proof
Let $(1)$ be rearranged as:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac y {y^2 e^y + x}$

Hence:
 * $(2): \quad \dfrac {\mathrm d x} {\mathrm d y} - \dfrac 1 y x = y e^y$

It can be seen that $(2)$ is a linear first order ODE in the form:
 * $\dfrac {\mathrm d x}{\mathrm d y} + P \left({y}\right) x = Q \left({y}\right)$

where:
 * $P \left({y}\right) = -\dfrac 1 y$
 * $Q \left({y}\right) = y e^y$

Thus:

Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
 * $\dfrac {\mathrm d} {\mathrm d y} \left({\dfrac x y}\right) = e^y$

and the general solution is:
 * $\dfrac x y = e^y + C$

or:
 * $x = y e^y + C y$