= Solution
For negative $\epsilon$ the fast arrows reverse: the outer branches become repelling and the middle branch attracting. On the middle branch the slow equation is $\dot x=x/(1-x^2)$, so the slow motion goes away from the <saddle equilibrium> at zero towards a fold.
At $(3,1)$, $F=3-9+1=-5$, so $\dot x>0$: the <trajectory> immediately escapes to the right. At $(1,-2)$, $F=1-1/3-2=-4/3$, so it also moves rapidly right and escapes. In each case the large-$x$ balance is $\dot x\sim x^3/(3|\epsilon|)$, with $y$ decreasing; it is not a bounded relaxation cycle.
At $(-1,-1/3)$, $F=-1$, so the initial fast motion is rightward. At nearly fixed $y=-1/3$ it approaches the attracting middle root of $x-x^3/3=1/3$, approximately $x=0.347$. It then moves slowly down/right along the middle branch towards $(1,-2/3)$, leaves that fold to the right, and eventually escapes. These are the small-negative-$\epsilon$ limiting descriptions; the fold locations and fast landing point receive small corrections. The right panel of the figure distinguishes this slow excursion from the two immediate escapes.
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