= Solution
A $C^1$ <function> $V$ near an equilibrium at the origin is a <Lyapunov function> if $V(0)=0$, $V(x)>0$ for $x\ne0$, and
$$
\dot V(x)=\nabla V(x)\cdot f(x)\leq0.
$$
The <First Lyapunov theorem> says these conditions imply stability. The second says that if the last inequality is strict away from the origin, then the origin is asymptotically stable.
To prove the first, fix a sufficiently small ball of radius $\varepsilon$. Positive definiteness and compactness give
$$
m=\min_{|x|=\varepsilon}V(x)>0.
$$
Continuity at zero gives $\delta>0$ such that $|x_0|<\delta$ implies $V(x_0)<m$. Since $V$ cannot increase along a trajectory, that trajectory cannot meet the sphere $|x|=\varepsilon$, where $V\geq m$. It therefore remains in the $\varepsilon$-ball for all forward time, which is stability.
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