$ (x^2y^3+2y)dx+(2x-2x^3y^3)dy=0$

$M=x^2y^3+2y\\ N=2x=2x^3y^2$

$\frac{\partial M}{\partial y}=3x^2y^2+2$

$\frac{\partial N}{\partial x}=2-6x^2y^2$

$\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$

The given Equation is not an exact differential equation.

The given equation is of the form $f_1(xy)ydx+f_2(xy)xdy=0$

$I.F=\frac{1}{Mx-Ny}$

$= \frac{1}{(x^2y^3+2x)x-(2x-2x^3y^2)y}$

$= \frac{1}{x^3y^3+2xy-(2xy-2x^3y^3)}$

$=\frac{1}{3x^3y^3}$

Multiplying the given equation by the integrating factor, we get

$(x^2y^3+2y)\frac{1}{3x^3y^3}dx+(2x-2x^3y^3)\frac{1}{3x^3y^3}=0$

$\left( \frac{1}{3x}+\frac{2}{3x^3y^2}\right)dx+\left( \frac{2}{3x^2y^3}-\frac{2}{3y}\right)dy=0$

$\frac{\partial M}{\partial y}=\frac{-4}{3x^3y}$

$\frac{\partial N}{\partial x}=\frac{-4}{3x^3y} $

The given Equation is an exact differential equation

Solution is given by

$\int Mdx+\int (Terms\ in\ N\ not\ containing\ x)dy=c$

$y=constant$

$\int \left( \frac{1}{3x}+\frac{2}{3x^3y^2}\right)dx+\int -\frac{2}{3y}=c$

$\frac{1}{3}\log x+ \frac{2}{3(-2)x^2y^2}-\frac{2}{3}\log y=c$

$\frac{1}{3}\log x- \frac{1}{3x^2y^2}-\frac{2}{3}\log y=c $