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 $a,b \in \mathbb R^+$