= Solution
Linearization about $X(t)$ gives
$$
\dot\Phi(t)=Df(X(t))\Phi(t),\qquad \Phi(0)=I.
$$
One <Floquet multiplier> of $\Phi(T)$ is $1$, corresponding to a phase shift along the orbit. Liouville's formula gives
$$
\det\Phi(T)=\exp\left(\int_0^T\nabla\cdot f(X(t))dt\right).
$$
In the plane this <determinant> is the other, transverse multiplier. The periodic orbit is transversely asymptotically stable when the displayed quantity is less than one, equivalently when the divergence <integral> is negative, and unstable when it is greater than one.
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