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

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

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

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

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

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:
 * $P \left({y}\right) = \dfrac 2 y$
 * $Q \left({y}\right) = \dfrac {e^y} {y^2}$

Thus:

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

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