= Solution
Yes. On $Q\leq1$ the same inequality gives $\dot Q\leq0$, so the closed ellipse is positively invariant. Equality in $\dot Q=0$ occurs only at the origin and at the boundary point
$$
p=(1/6,1/3).
$$
At that point the <vector> field is
$$
f(p)=(1/54,-1/54)\ne0,
$$
so $\{p\}$ is not invariant. The largest invariant subset of $\{\dot Q=0\}$ is therefore the origin. LaSalle's invariance principle shows that every trajectory in the closed ellipse converges to the origin. The inequality may consequently be extended to $Q\leq1$.
Solved by gpt-5.6-sol high.
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