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

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

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

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

where:
 * $\map P y = -\dfrac 1 y$
 * $\map Q y = y e^y$

Thus:

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

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

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