= Solution
On the attracting right branch, $x>1$ and $\dot y=-x<0$, so the motion goes from $(2,2/3)$ towards the fold $(1,-2/3)$. At essentially fixed $y=-2/3$ it jumps to the other attracting root $x=-2$. The left branch then carries it to $(-1,2/3)$, where it jumps right to $x=2$. This gives the clockwise limiting <relaxation oscillation> shown in the preceding figure.
Along a slow branch $y=h(x)$, differentiation gives $(x^2-1)\dot x=-x$. The right-branch time is
$$
T_R=\int_2^1\frac{1-x^2}{x}\,dx=\left[\log x-\frac{x^2}{2}\right]_2^1=\frac32-\log2.
$$
Reflection symmetry gives the same time on the left. The fast jumps take a time tending to zero as $\epsilon\to0^+$, as do the fold-transition corrections at leading order. Hence
$$
\boxed{T\sim2T_R=3-2\log2.}
$$
This is a leading asymptotic period, not an exact finite-$\epsilon$ equality.
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