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

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

$$v=\frac{y}{x}$$

$$y=v x$$

$$y'=x v'+v$$

$$(x-v x)(x v'+v)=(x+v x)$$

$$v x - v^2 x+x^2 v' - v x^2 v' = x + v x$$

$$x^2(v' - v v')=x(1+v^2)$$

$$x v'(1-v)=(1+v^2)$$

$$\frac{1-v}{1+v^2}dv=\frac{dx}{x}$$