= Solution
For $\dot x=f(x)$, a fixed point $x^*$ satisfies $f(x^*)=0$. It is <Lyapunov stability>[Lyapunov stable] when, for every $\varepsilon>0$, there is $\delta>0$ such that
$$
|x(0)-x^*|<\delta
\quad\Longrightarrow\quad
|x(t)-x^*|<\varepsilon\quad(t\geq0).
$$
A <Lyapunov function> on a neighbourhood of $x^*$ is a continuously differentiable function $V$ such that
$$
V(x^*)=0,
\qquad V(x)>0\quad(x\ne x^*),
\qquad
\dot V(x)=\nabla V(x)\cdot f(x)\leq0.
$$
The <First Lyapunov theorem> states that such a function makes $x^*$ Lyapunov stable. If $\dot V<0$ away from $x^*$, then $x^*$ is locally asymptotically stable.
To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^*|=\varepsilon$, compactness and positive definiteness give
$$
m=\min_{|x-x^*|=\varepsilon}V(x)>0.
$$
Continuity at $x^*$ supplies $\delta<\varepsilon$ such that $|x-x^*|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.
If $\dot V<0$ away from $x^*$, take a sufficiently small compact sublevel set of $V$. The <LaSalle invariance principle> says that every trajectory in it approaches the largest invariant subset of $\{\dot V=0\}$, which is just $\{x^*\}$. Hence the equilibrium is also <asymptotic stability>[asymptotically stable].
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