Answer
See answer below
Work Step by Step
$y(x^2-2xy)dx-x^3dy=0$ ___(1)
Multiply (1) by $I=y^{-2}e^{-\frac{x}{y}}$
$y^{-2}e^{-\frac{x}{y}}(y(x^2-2xy)dx-x^3dy)=y^{-1}e^{-\frac{x}{y}}(x^2-2xy)dx-x^3y^{-2}e^{-\frac{x}{y}}dy$ ___(2)
Here, $M_{1}(x,y)=y^{-2}e^{-\frac{x}{y}}(y(x^2-2xy)$
$N_{1}(x,y)=-x^3y^{-2}e^{-\frac{x}{y}}$
$(M_{1})_y=\frac{x^2e^{-\frac{x}{y}}(x-3y)}{y^3}$
$(N_{1})_x=\frac{x^2e^{-\frac{x}{y}}(x-3y)}{y^3}$
$(M_{1})_y=(N_{1})_x$
$\Rightarrow $ (2) is Exact
$I=y^{-2}e^{-\frac{x}{y}}$ is integrating factor of (1).
