The asymptotic stability if the time derivative of Lyapunov function is zero

Question

The dynamic equation of motion of the system is \begin{align} \ddot{\theta} + \frac{g}{l}\sin\theta = 0 \end{align} We should consider first the total energy function as a Lyapunov function candidate, hence: \begin{align} V(\dot{\theta},\theta) &= \frac{1}{2} ml^2 \dot{\theta}^2 + mgl(1-\cos\theta) \tag{1} \end{align} The Eq.(1) is not positive definite function because $V(\dot{\theta},\theta) = 0$ at $\theta \neq 0$ (i.e. at $\theta = 2n\pi, \ n\in \mathbb{Z}$). If the domain of $\theta$ in $(-2\pi,2\pi)$, then the Eq.(1) is indeed positive definite. The time-derivative of $V(\dot{\theta},\theta)$, hence: \begin{align} \dot{V}(\theta,\dot{\theta}) &= \frac{\partial V(\theta,\dot{\theta})}{\partial \dot{\theta}} \frac{\partial \dot{\theta}}{ \partial t} + \frac{\partial V(\theta,\dot{\theta})}{\partial \theta} \frac{\partial \theta}{ \partial t} \notag \\ &= (ml^2 \dot{\theta}) \ddot{\theta} + (mgl\sin\theta) \dot{\theta} \notag \\ &= (ml^2 \dot{\theta}) (-\frac{g}{l}\sin\theta) + (mgl\sin\theta) \dot{\theta} \notag \\ &= -mgl \dot{\theta}\sin\theta + mgl\dot{\theta}\sin\theta \notag \\ &= 0 \end{align} Since $\dot{V}(\theta,\dot{\theta}) \leq 0$, the origin is stable. Is it possible to investigate the asymptotic stability if $\dot{V}(\theta,\dot{\theta})=0$. As far I as I know, we can use the invariance principle (i.e. LaSalle's theorem), but I can't see how to utilize this theorem in case $\dot{V}(\theta,\dot{\theta})=0$. Any suggestions?!