= Solution
With $\nu<0$ and small negative $\mu$, the returning trajectory enters to the left of the bottleneck, crosses the region around $x=0$ slowly, and exits to the right. Its height contracts towards $y=0$ throughout because $\dot y=-\lambda y$. The local horizontal arrows point right everywhere. There is no <equilibrium point>; the schematic global excursion then returns the trajectory to the incoming section, giving the stable <limit cycle> shown in the negative-$\mu$, negative-$\nu$ panel.
For fixed $h>0$, the complete local bottleneck crossing takes
$$
\int_{-h}^h\frac{dx}{x^2-\mu}
=\frac2{\sqrt{-\mu}}\arctan\frac h{\sqrt{-\mu}}
=\frac\pi{\sqrt{-\mu}}-\frac2h+O(-\mu).
$$
The actual entry coordinate is near a fixed negative $\nu$. Replacing $-h$ by that entry removes only a bounded time as $\mu\to0^-$, and the global excursion also takes bounded time. Therefore
$$
\boxed{T\sim\frac\pi{\sqrt{-\mu}}.}
$$
This is the <inverse-square-root period law near a saddle-node bottleneck>. The coefficient $\pi$ assumes fixed negative $\nu$, or more generally $-\nu/\sqrt{-\mu}\to\infty$. In a joint approach to the codimension-two point, the actual local time is $[\arctan(h/k)-\arctan(\nu/k)]/k$, $k=\sqrt{-\mu}$. If $\nu/k\to-C$, the coefficient is $\pi/2+\arctan C$ instead; at entry $\nu=0$ it is $\pi/2$. The fixed-negative-$\nu$ law is not uniform over all joint parameter paths.
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