First Order ODE/dx = (y over (1 - x^2 y^2)) dx + (x over (1 - x^2 y^2)) dy

Theorem
The first order ordinary differential equation:


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

is an exact differential equation with solution:


 * $\ln \left({\dfrac {1 + x y} {1 - x y} }\right) - 2 x = C$

This can also be presented as:
 * $\dfrac {\mathrm d y} {\mathrm d x} = -\dfrac {\dfrac y {1 - x^2 y^2} - 1} {\dfrac x {1 - x^2 y^2}}$

Proof
First express $(1)$ in the form:
 * $(2): \quad \left({\dfrac y {1 - x^2 y^2} - 1}\right) + \left({\dfrac x {1 - x^2 y^2}}\right) \dfrac {\mathrm d y} {\mathrm d x}$

Let:
 * $M \left({x, y}\right) = \dfrac y {1 - x^2 y^2} - 1$
 * $N \left({x, y}\right) = \dfrac x {1 - x^2 y^2}$

Then:

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $f \left({x, y}\right) = C$

where:

Hence:

Substituting $z = x y$ and obtain:
 * $\dfrac {\mathrm d z} {\mathrm d x} = y$

This gives:

and:

Thus:
 * $f \left({x, y}\right) = \dfrac 1 2 \ln \left({\dfrac {1 + x y} {1 - x y} }\right) - x$

and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:
 * $\ln \left({\dfrac {1 + x y} {1 - x y} }\right) - 2 x = C$