In a textbook I have seen a stability result known as Dirichlet's theorem: let $$\dot{x} = f(x)$$ be $C^{r}$ a vector field on $\mathbb{R}^{n}$ where $r \geq 1$, with a fixed point at $x = \bar{x}$. Let $I(x)$ be first integral of the vector field defined in a neighbourhood of $x = \bar{x}$ such that $x = \bar{x}$ is a nondegenerate minimum of $I(x)$. Then $x = \bar{x}$ is Lyapunov stable.
My proof: since $I(x)$ is the first integral, then $\dot{I}(x) = 0$. Since $\bar{x}$ is the nondegenerate minimum of $I(x)$, we have $I(x) > I(\bar{x})$ for all $x \neq \bar{x}$, in a neighbourhood of $\bar{x}$. Define $\tilde{I}(x) = I(x) - I(\bar{x})$. Then $\tilde{I}(\bar{x}) = 0$ and $\tilde{I}(x) >0$ for $x \neq \bar{x}$, in an neighbourhood of $\bar{x}$. Since $\dot{\tilde{I}} = \dot{I}=0$, we may define $\tilde{I}$ to be the Lyapunov function and by the above criteria $\bar{x}$ is indeed stable.
Is this correct?