Solving the differential equation $(x^2-y^2)y' - 2xy = 0$. I am trying to solve the equation
$$
(x^2-y^2)y' - 2xy = 0.
$$
I have rearranged to get 
$$
y' = f(x,y)
$$ 
where 
$$
f(x,y) = \frac{2xy}{x^2-y^2}.
$$
From here I tried to use a trick that I learned from class where we set $x=\frac{y}{x}$ and write
$$
\frac{z'}{f(1,z)-z} = \frac{1}{x}.
$$
After doing a calculation I have that 
$$
\frac{z'}{f(1,z)-z} = z'\frac{(1-z^2)}{(z^3+z)}.
$$
From here I would like to use the fact that 
$$
z'\frac{(1-z^2)}{(z^3+z)} = \frac{1}{x}
$$
and integrate both sides. However, I'm stuck at integrating the left-hand side. 
How can I integrate the left-hand side, or is there is a better way to solve this equation? Thanks.
 A: Correction (after missing a sign:) As kobe pointed out, the original DE is $$
(x^2-y^2)y'-2xy=0,
$$ 
which as equation for a vector field reads
$$
(x^2-y^2)\,dy-2xy\,dx=0\iff Im(\bar z^2\,dz)=0\text{ with } z=x+iy.
$$ 
From the complex interpretation it is directly visible that this is not integrable, for that it would have to be an expression $Im(f(z)\,dz)$. But since $\bar z=|z|^2/z$, an integrating factor presents itself as $|z|^{-4}=(x^2+y^2)^{-2}$ which repairs this deficiency. Then
$$
\frac{(x^2-y^2)\,dy-2xy\,dx}{(x^2+y^2)^2}=Im(z^{-2}\,dz)=-d(Im(z^{-1}))
$$ 
which implies that all solution trajectories lie on the curves 
$$
-Im(z^{-1})=\frac{y}{x^2+y^2}=C\quad(\in \Bbb R)
$$
which can be solved as
$$
y^2-2y(2C)^{-1}+(2C)^{-2}=(2C)^{-2}-x^2\\
y=(2C)^{-1}\pm\sqrt{(2C)^{-2}-x^2}
$$
giving the solutions $y\equiv0$ and $y=D\pm\sqrt{D^2-x^2}$ on the interval $x\in[-D,D]$.


Variant: If one reads too fast that the sign before $2xy$ were reversed, then
$$
(x^2-y^2)dy+2xydx=0\quad (\iff Im(z^2·dz)=0)
$$
is an exact differential equation with conserved quantity or first integral $x^2y-\frac13y^3$ ($=Im(\frac13 z^3)$) so that the solutions lie on the curves
$$
y(y^2-3x^2)=const.
$$
A: Hopefully it was just a typo the sub you have. 
Anyway, using $y = vx$ we find 
$$
y' = \frac{2x^2v}{x^2 - x^2v^2} = \frac{v}{1-v^2}
$$
Then we have
$$
v + xv' = \frac{2v}{1-v^2}
$$
I will spend more time to find where exactly you went wrong because it looks a little strange (to me anyway) 
$$
xv' = v\frac{v^2-1 + 2}{1-v^2} = -v\frac{v^2 +1}{v^2-1}
$$
This leads to
$$
\int \frac{v^2}{v\left(v^2+ 1\right)} -\frac{1}{v\left(v^2+ 1\right)}dv 
$$
The first integral is straight forward after reducing the fraction. The second makes a little more work
$$
\int \frac{1}{v\left(v^2+ 1\right)} dv 
$$
Let $v = \tan t$ then we have 
$$
\int \frac{1}{\tan t \left(\tan^2 t +1\right)} \sec^2t dt = \int \frac{\sec^2 t}{\tan t \sec^2 t} dt = \int \frac{\cos t}{\sin t} dt
$$
Then replace all the subs you will be done. 
A: You don't need to do any of that. This is an exact differential equation. The solution has the form $$F(x, y) = C$$ where $$\frac{\partial F}{\partial x} = 2xy$$ and $$\frac{\partial F}{\partial y} = x^2 - y^2$$
