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

My work:

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

$$\frac{dy}{dx}=x+y-\frac{xy}{x-y}=x+y\left[1-\frac{x}{x-y}\right]$$

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

I tried to make it first order linear differential equation and then I also tried to make it exact, but I still had trouble.

I tried substitution : $Z=1/(x-y)$