= Solution
A <Lyapunov function> near an equilibrium is a continuously differentiable function with $V(0)=0$, $V(x)>0$ for nonzero nearby $x$, and $\dot V=\nabla V\cdot f\leq0$ along trajectories there. Strict negativity away from the equilibrium gives local <asymptotic stability>. <Lyapunov stability> itself means that every prescribed small neighborhood contains a smaller neighborhood whose forward trajectories never leave the prescribed one.
For the proposed quadratic form, direct differentiation gives
$$
\dot V=-2[(x-y)^2+(\beta-1)y^2]+4x^2y-2x^3+(2\beta-8)xy^2.
$$
When $\beta>1$, the quadratic part is negative definite, so it dominates the cubic remainder in a sufficiently small ball; $V$ is then a <strict Lyapunov function>. If $0<\beta<1$, take $x=y$ to make the leading derivative positive. At $\beta=1$, that same line gives $\dot V=-4x^3$, positive on its negative half. Values $\beta\leq0$ fail positive definiteness of $V$. Therefore
$$
\boxed{\beta>1.}
$$
For $\beta=2$ the full derivative factors, with no small-amplitude truncation:
$$
\dot V=-2(1+x)[(x-y)^2+y^2],\qquad V=x^2+2y^2.
$$
Every closed sublevel set $V\leq c<1$ lies in $x>-1$, is forward invariant, and has strictly decreasing $V$ away from the origin. Compactness and the <Lyapunov stability> argument therefore show that \b[the ellipse $x^2+2y^2<1$ belongs to the origin's basin of attraction]. Boundary points with $V=1$ other than $(-1,0)$ also have negative derivative and enter the ellipse. The exception $(-1,0)$ is another <fixed point>, so the largest such sublevel ellipse cannot be included wholesale in the basin. This is a <basin estimate from a Lyapunov sublevel set>, not a claim that no additional points outside the ellipse converge to the origin.
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