 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 isReflection symmetry gives the same time on the left. The fast jumps take a time tending to zero as $ϵ\to 0^{+}$, as do the fold-transition corrections at leading order. HenceThis is a leading asymptotic period, not an exact finite-$ϵ$ equality.