How do I solve $y'=\frac{x-y}{x+y}$? This is what I have so far:

$y'=\frac{x-y}{x+y}$

$ y'=\frac{x}{x+y}-\frac{y}{x+y}$

$ y'=\frac{1}{1+\frac{y}{x}}-\frac{1}{\frac{x}{y}+1} $

Make a substitution here:

$v=\frac{y}{x}, y=vx, y'=v'x+v$

$ v'x+v = \frac{1}{1+v}-\frac{1}{\frac{1}{v}+1}$

Do I just simplify the right side here, and then separate variables?  When I tried this, I got:

$\int \frac{1+v}{1-2v-v^2}dv = lnx+C_0$

I didn't know how to solve the integral here.

Any ideas?
You are complicating it. Sum $1$ to both sides to get $$y'+1=\frac{x-y}{x+y}+1=\frac{2x}{x+y}.$$ Now, because $(x+y)'=y'+1$, we get $$(x+y)'=\frac{2x}{x+y}.$$
Familiar, right? Write $u=y+x$, so $u'=2x/u \implies u^2/2=x^2+A$. We finally have $$y=-x\pm\sqrt{2x^2+B}.$$
You can solve this integral by using the partial fractions technique.
But tell me this. That simplification is correct? You got this result below:
$$\int \frac{1+v}{1-2v-v^2}dv$$

I've got this one: 

$$\int \frac{1-v}{1-v}dv$$

That is 1 if v is different than 1, right?

Am I off the track here?

You got $$ \frac{1}{1-v}- \frac{1}{\frac{1}{v}-1} = K$$

$$\therefore  K = \frac{1}{1-v} -\frac{1}{\frac{1-v}{v}} $$

$$  \therefore K = \frac{1}{1-v} - \frac{v}{1-v}  $$

$$ \therefore K = \frac{1-v}{1-v}=1$$

So v can't be 0 and can't be 1 neither.