Solving the exact differential equation y'=(x-y)/(x+y) I need to solve the following exact equation:
$y' = \frac{(x-y)}{(x+y)}$
I've been taught to put those in the form $M(x, y)dx + N(x, y)dy = 0$ and to make sure ${dM}\over{dy}$ = ${dN}\over{dx}$
So I transform the equation into
$\frac{1}{(x+y)} dx + \frac{1}{(y-x)}dy=0$
However, $\frac{dM}{dy}=\frac{-1}{(y+x)^2}$ and $\frac{dN}{dx}=\frac{1}{(x-y)^2}$
They are not equal? I'm not supposed to use an integrating factor either.. so I'm a bit at a loss :(
The answer is supposed to be $x^2 -2xy -y^2 = c$
How do I get there?
Thanks a lot in advance :)

I do not know if you are obliged to go through these steps; so forgive me if what I write is not an answer to your post.
The differential equation simplifies a lot if you define $x+y=z$; so $y=z-x$, $y'=z'-1$ and the rhs write $\frac {2x-z}{z}=\frac {2x}{z}-1$. Simplifying then leads to $$z z'=2x$$which, by integration of both sides, leads to $$z^2=2x^2+C$$ Back to the definition of $z$,$$(x+y)^2=2x^2+C$$ which, after development, leads to $$x^2 -2xy -y^2 = C$$
