= Solution
The required properties are those of a <strict Lyapunov function>; the domain of stability is the set of initial states whose forward trajectories converge to the fixed point.
For $V=(x^2+y^2)/2$,
$$
\dot V=(x^2+y^2)(-2+x^2+4y^2).
$$
Thus $\dot V<0$ when $V<1/4$, proving that disc lies in the basin. On $V=1$, $\dot V=6y^2\ge0$ (with the tangency points checked from the <vector> field), so trajectories outside cannot enter the basin through that boundary. Hence the basin is contained in $V\le1$.
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