<s>[INST] Solve $\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$ 
Solve $\dfrac{xdx-ydy}{xdy-ydx}=\sqrt{\dfrac{1+x^2-y^2}{x^2-y^2}}$

I tried doing this-
$\sqrt{x^2-y^2}d(x^2-y^2)=x^2\sqrt{1+x^2-y^2}d(\frac{x}{y})$ but it does not get better from here.
I even tried putting $\sqrt{x^2-y^2}$ as $u$ but that does not simplify either.
 [/INST] I would suggest to you a more intricate change of variables:
\begin{align}
x &= z \cosh{s} = z \,\left(\frac{e^{s} + e^{-s}}{2}\right)\\
y &= z \sinh{s} = z \left(\frac{e^{s} - e^{-s}}{2}\right)
\end{align}
Then 
\begin{align}
dx = \cosh(s) \, dz + z\,\sinh(s)\, ds \\ 
dy = \sinh(s) \, dz + z\,\cosh(s)\, ds  
\end{align}
as well as
$$x^2-y^2 = z^2\big(\cosh^2(s) - \sinh^2(s)\big) = z^2$$ which means that
$$z = \sqrt{x^2 - y^2}$$
Now
\begin{align}
y\,dx =   z \sinh(s) \cosh(s) \, dz + z^2\,\sinh^2(s)\, ds \\ 
x\,dy = z \cosh(s) \sinh(s) \, dz + z^2\,\cosh^2(s)\, ds  
\end{align}
which leads to the difference
\begin{align}
x\,dy - y\,dx &= z \cosh(s) \sinh(s) \, dz + z^2\,\cosh^2(s)\, ds\\
& -   z \sinh(s) \cosh(s) \, dz - z^2\,\sinh^2(s)\, ds \\
&=   z^2\,\cosh^2(s)\, ds -  z^2\,\sinh^2(s)\, ds \\ 
&= z^2\big(\cosh^2(s) - \sinh^2(s)\big) \, ds\\
&= z^2\, ds
\end{align}
Moreover, 
\begin{align}
x\,dx - y\,dy &= \frac{1}{2}d \left(x^2-y^2\right)\\ 
&= \frac{1}{2} d(z^2)\\ 
&= z\,dz
\end{align}
Consequently
\begin{align}\frac{x\,dx - y\,dy}{ x\,dy - y\,dx } &= \frac{z\, dz}{z^2 \, ds}\\ &= \frac{1}{z} \frac{dz}{ds}\end{align}
and finaly the equations becomes
Moreover, 
\begin{align}
 \frac{1}{z} \frac{dz}{ds} = \frac{x\,dx - y\,dy}{ x\,dy - y\,dx } = \sqrt{\frac{1 + x^2-y^2}{x^2-y^2}} = \sqrt{\frac{1 + z^2}{z^2}} = \frac{\sqrt{1+z^2}}{z}
\end{align} 
multiply both sides by $z$ and obtain the simple differential equation
$$ \frac{dz}{ds} = \sqrt{1+z^2}.$$
I think you can take it from here.
</s>