 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 $V=(x^2+y^2)/2$,Thus $\dot{V}<0$ when $V<1/4$, proving that disc lies in the basin. On $V=1$, $\dot{V}=6y^2\ge 0$ (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 $V\le 1$.