= Solution
The <nullcline> is $\boxed{y=h(x)=x^3/3-x}$. Write $F=x-x^3/3+y$. Away from it at an order-one distance, $F$ is order one on the indicated finite phase-space scales. Thus $\dot x=O(\epsilon^{-1})$ while $\dot y=-x=O(1)$: the <trajectory> changes $x$ on the <fast time in a singular perturbation> $t/\epsilon$, with only an $O(\epsilon)$ change in $y$. Equivalently $dy/dx=-\epsilon x/F=O(\epsilon)$, which explains the nearly horizontal motion.
For positive $\epsilon$, $F_x=1-x^2$ makes the branches $|x|>1$ fast attracting and the middle branch $|x|<1$ fast repelling. The arrows point right where $y>h(x)$ and left where $y<h(x)$, carrying a generic off-<nullcline> point towards an attracting branch. The reduced description changes near a fold, where $F_x=0$; the rapid/slow distinction there needs a separate transition layer.
\Image[/past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2009-relaxation-phase-plane.png]
{title=Cubic nullcline, limiting relaxation cycle, and reversed fast stability}
The left panel shows positive $\epsilon$; the right panel shows negative $\epsilon$, including the three specified starting points. Horizontal arrows indicate the leading fast motion.
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