= Solution
Put $y=\dot x$. The first-order system is
$$
\dot x=y,
\qquad
\dot y=-2x(1-x^2)^2-\mu y.
$$
At an <equilibrium point of a dynamical system>, $y=0$ and $x(1-x^2)^2=0$, so the three fixed points are
$$
\boxed{(0,0),\ (1,0),\ (-1,0).}
$$
Define
$$
U(x)=x^2-x^4+\frac{x^6}{3},
\qquad
E(x,y)=\frac12y^2+U(x).
$$
Then $U'(x)=2x(1-x^2)^2=-F(x)$ and
$$
\dot E
=y\dot y+U'(x)\dot x
=-\mu y^2\leq0.
$$
Moreover,
$$
U(x)=x^2\left(1-x^2+\frac{x^4}{3}\right)>0
\quad(x\ne0),
$$
so $E$ is positive definite at the origin. It is therefore a <damped mechanical energy as a Lyapunov function>, and the <First Lyapunov theorem> proves that the origin is Lyapunov stable.
Since $U(\pm1)=1/3$, choose a compact energy sublevel $\{E\leq c\}$ with $0<c<1/3$. In this set, $\dot E=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm1$. Thus the largest invariant subset of $\{\dot E=0\}$ is the origin. The <LaSalle invariance principle> proves that the origin is asymptotically stable.
Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its <omega-limit set> inside
$$
\{(x,0):F(x)=0\}
=\{(-1,0),(0,0),(1,0)\}.
$$
An omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.
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