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

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

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

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

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

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

where:
 * $\map P y = \dfrac 2 y$
 * $\map Q y = \dfrac {e^y} {y^2}$

Thus:

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

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