= Solution
The only <equilibrium point> is $(0,0)$. Its <Jacobian matrix> is $\begin{pmatrix}-\mu&1\\-1&0\end{pmatrix}$, with <characteristic polynomial> $\lambda^2+\mu\lambda+1$. For $\mu>0$ both <eigenvalues> have negative real parts, so the origin is locally <asymptotically stable>.
Use the <Lyapunov function> $V=(x^2+y^2)/2$. Along a solution,
$$
\dot V=x\dot x+y\dot y=\mu x^2(x^2/3-1).
$$
A <Lyapunov function> is positive away from the equilibrium and nonincreasing along trajectories; its sufficiently small sublevel sets prove stability. <Asymptotic stability> additionally means nearby trajectories converge to the equilibrium. The <basin of attraction> is the set of initial states whose forward solutions exist for all positive time and converge to it. Every closed disc of radius at most $\sqrt3$ is positively invariant, since $\dot V\leq0$ there. The <LaSalle invariance principle> says that a solution in a compact positively invariant set where $\dot V\leq0$ approaches the largest invariant subset of $\{\dot V=0\}$. Here that subset is only the origin: on $x=0$, remaining on the line would require $y=0$; the isolated points $x=\pm\sqrt3,y=0$ on the outer disc do not remain there because $\dot y=-x\ne0$. Hence \b[the closed disc $x^2+y^2\leq3$ lies in the <basin of attraction>.]
This estimate does not assert that the basin is exactly that disc, or the whole plane. For example, choose $R$ so large that $R^2\geq6(\mu+1)/\mu$ and $R^2>3(\mu+2)/\mu$. The region $x\geq R,\ y\geq-x$ is positively invariant: on $x=R$ one has $\dot x>0$, and on $y=-x$ one has $(x+y)^{\boldsymbol\cdot}=\mu x^3/3-(\mu+2)x>0$. Within it $\dot x\geq\mu x^3/6$, which forces finite-time escape. Thus the attraction is genuinely local. The <Lyapunov function> gives a rigorous sufficient domain for attraction without identifying its entire boundary.
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