First, Note that $x,y \neq \pm1$ and $\ln C = \frac 12 \ln C^2$. Denote $C^2=k >0$. So the equation becomes
$\frac 12 (\ln |y+1| -\ln |y-1| + \ln k ) = \frac 12 (\ln |x+1| - \ln |x-1| )$
Cancelling $\frac 12$ and using basic logarithm identities, we get,
$ \ln \left ({ k \left |\frac {y+1}{y-1}\right |}\right ) = \ln \left ( \left | \frac {x+1}{x-1} \right | \right ) $
Now, we get, $k\left |\frac {y+1}{y-1}\right | = \left | \frac {x+1}{x-1} \right |$
Now simplify it. Best way is to break into different cases.
Case 1: $ |y| > 1 $ and $|x| >1$
Gives $k \frac {y+1}{y-1} = \frac {x+1}{x-1} $ Solve for $y$ in term of $x$ or vice versa as desired.
Other cases can be dealt with similarily.
Note: Seems like you are starting to learn about logarithms, so I'll write the facts I used;
$\ln a + \ln b = \ln ab$ $\ln a - \ln b = \ln \frac ab$ $\ln a = \ln b \iff a=b$
all of these holds for $a,b \in \mathbb R^+$