Solution:
Twice, at $(1,0)$ and in the third quadrant.

First $x \geq 1$ or $x<-1$. Solving $x$ in terms of $y$, we get $x=\frac{1+y^{2}}{1-y^{2}}$. The graph of this equation has asymptotes $y=1$ and $y=-1$. For $x \geq 1$ in this graph, $-1<y<1$ and this portion of the graph does not intersect by $y=x$. For $x<-1$, we have $y>1$ or $y<-1$ with $x=-1$ as asymptote. $y=x$ intersects this portion of the graph once in the third quadrant.