Answer
$I(x,y)=x^2y^4$
Work Step by Step
We are given
$(y^{-1}-x^{-1})dx+(xy^{-2}-2y^{-1})dy=0$
Multiplying equation (1) by $I(x)=x^ry^s$
$x^ry^s(y^{-1}-x^{-1})dx+x^ry^s(xy^{-2}-2y^{-1})dy=0$
Here, $M(x,y)=x^ry^s(y^{-1}-x^{-1})$
$N(x,y)=x^ry^s(xy^{-2}-2y^{-1})$
$\frac{\partial M}{\partial y}=x^{r-1}y^{s-2}((s-1)x-sy)(rx-2ry+x)$
$\frac{\partial N}{\partial x}=x^{r-1}s^{r-2}$
The given diffirential equation is exact
$\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}$
$x^{r-1}y^{s-2}((s-1)x-sy)=x^{r-1}y^{s-2}(rx-2ry+x)$
$(s-1)x-sy=(r+1)x-2ry$
$s=2r$
Hence here,
$I(x,y)=x^2y^4$
