= Solution
For $|y|>1$, the separated integral is
$$
\frac12\log\left|\frac{1+y}{1-y}\right|=x+C.
$$
Inverting gives
$$
\boxed{y(x)=\coth(x+C).}
$$
The upper branch has $x+C>0$, where $y>1$ decreases towards $1$ as $x\to\infty$. The lower branch has $x+C<0$, where $y<-1$ decreases to $-\infty$ as $x\uparrow-C$. These are separate maximal intervals: the pole is not part of a solution. For the domain $x\ge0$, a lower-branch initial value below $-1$ necessarily reaches that pole in finite positive $x$. The <direction field> sketch above displays both exterior branches and a central <hyperbolic tangent> branch.
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