If you must solve this analytically, you may try this:
$$
\left(
\begin{array}{c}
x_{n+1}\\
x_n
\end{array}
\right)=\left(
\begin{array}{cc}
-y-1/y&-1\\
1&0
\end{array}
\right)\left(
\begin{array}{c}
x_n\\
x_{n-1}
\end{array}
\right).
$$
This is of the form
$$
\mathbf{x}_n=A\mathbf{x}_{n-1},
$$
whose solution reads
$$
\mathbf{x}_n=A^{n-1}\mathbf{x}_1.
$$
As for $A^{n-1}$, you need to diagonalize $A$ at first, i.e., $A=P\Lambda P^{-1}$, where $\Lambda$ is a diagonal matrix. In this way, $A^{n-1}=P\Lambda^{n-1}P^{-1}$.
You need to leave $x_1$ undetermined, and set $x_2$ as per your last constraint. After you obtain the general solution to $x_n$ and $x_{n-1}$ (obviously, they depends on the value of $x_1$), you determine the value of $x_1$ by using your second constraint.