(x-y)y'=(x+y)
I've tried v=y/x and v=x-y, and both times it has just ended up as a mess.
the solution Mathematica is pumping out seems to indicate that v=y/x was the right way to go.
you can certainly separate the equation that way
$$(x-y)y'=(x+y)$$
$$v=\frac{y}{x}$$
$$y=v x$$
$$y'=x v'+v$$
$$(x-v x)(x v'+v)=(x+v x)$$
$$v x - v^2 x+x^2 v' - v x^2 v' = x + v x$$
$$x^2(v' - v v')=x(1+v^2)$$
$$x v'(1-v)=(1+v^2)$$
$$\frac{1-v}{1+v^2}dv=\frac{dx}{x}$$