$$\frac {dy}{dx}=\frac{x^2-xy-y^2}{x-y}$$
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 : $Z=1/(x-y)$
On
Hint:
$\dfrac{dy}{dx}=\dfrac{x^2-xy-y^2}{x-y}$
$(y-x)\dfrac{dy}{dx}=y^2+xy-x^2$
This belongs to an Abel equation of the second kind.
Let $u=y-x$ ,
Then $y=u+x$
$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$
$\therefore u\left(\dfrac{du}{dx}+1\right)=(u+x)^2+x(u+x)-x^2$
$u\dfrac{du}{dx}+u=u^2+2xu+x^2+xu+x^2-x^2$
$u\dfrac{du}{dx}=u^2+(3x-1)u+x^2$
Let $u=e^xv$ ,
Then $\dfrac{du}{dx}=e^x\dfrac{dv}{dx}+e^xv$
$\therefore e^xv\left(e^x\dfrac{dv}{dx}+e^xv\right)=e^{2x}v^2+(3x-1)e^x+x^2$
$e^{2x}v\dfrac{dv}{dx}+e^{2x}v^2=e^{2x}v^2+(3x-1)e^x+x^2$
$e^{2x}v\dfrac{dv}{dx}=(3x-1)e^x+x^2$
$v\dfrac{dv}{dx}=(3x-1)e^{-x}+x^2e^{-2x}$
It doesn't seem to have closed-form solutions.