$ x\frac{dy}{dx}=y-\frac{y^2}{x}\\ x\frac{dy}{dx}=\frac{xy-y^2}{x}$

$xdy=\frac{xy-y^2}{x}dx\\ x^2dy=(xy-y^2)dx$

$(xy-y^2)dx-x^2dy=0\\ M=xy-y^2 \\ N=-x^2$

$\frac{\partial M}{\partial y}=x-2y$

$ \frac{\partial N}{\partial x}=-2x$

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

The given Equation is not an exact differential equation.

M and N are homogenous functions of degree 3

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

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

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

$= \frac{1}{-yx^2}$

Multiplying the given equation by the integrating factor, we get

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

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

$ \frac{\partial M}{\partial y}=\frac{1}{y^2}$

$ \frac{\partial N}{\partial x}=\frac{1}{y^2} $

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}{y}+\frac{1}{x} \right)dx+\int 0dy=c$

$-\frac{x}{y}+\log x=c $