Do not solve, (x-y) + (y+x)y' = 0, via integrating factor. It would not work.

Instead solve for y,

y' = (y-x)/(y+x)

So the equation is of the form,

y' = h(x,y)

Where h(x,y) is a homosexual function i.e. h(ax,ay) = a*h(x,y).

Thus, use the substitution z = y/x to solve this equation.