Stability of critical points in a ODE and extremals of the constant of motion I'm doing some quality studies of some simple ODEs and I wondered how to make calculation shorter.
Suppose we have a planar differential system (Autonomous) \begin{cases}\dot{x} &= f(x,y)\\ \dot{y} &=g(x,y) \end{cases} defined over all $\mathbb{R}^2$. 
Suppose that exists a smooth non constant constant of motion $V\colon \mathbb{R}^2 \to \mathbb{R}$, such that $X_1 \in \mathbb{R}^2$ is an isolated local minimum (or local maxima, I don't think it is different) and so a critical point for the system.
Can I conclude that $X_1$ is a stable point for the system because it has a neighborhood full of cycles around it? More explicitly

thanks to the existence of the costant of motion $V$, exists a neighborhood of $X_1$ in which all of the orbits are cycles around the point. $ V $ is locally a paraboloid. These cycles are clearly positive and negative invariant sets, and their inner bounded regions (jordan theorem) form a fundamental system of neighborhoods for $X_1$. So I can conclude that the point is either positive stable and negative stable. 

But I don't think a point can be at the same time positive and negative stable. Can someone clarify this situation?
NB by positive stable point I mean a point with a fundamental system of neighborhood which are positive invariants. Negative stable point is analogue.
 A: It is possible for a point to be both  positive stable and negative stable, according to your definition. For example, the trajectories of the system $\dot x = -y$, $\dot y=x$  are circles centered at  $(0,0)$. Therefore, $(0,0)$ has a basis of neighborhoods which are invariant both forward in time and backward in time.  This is in fact a concrete  example of the kind of  system you consider, with $V(x,y)=x^2+y^2$. 
Your proof needs some work though. You say "$V$ is locally a paraboloid", which is both vague and untrue. The "paraboloid" probably refers to the 2nd order Taylor expansion of $V$, which indeed gives a paraboloid if the Hessian matrix of $V$ is positive definite or negative definite. But it could be semidefinite; even identically zero. The structure of level sets of $V$ can be a bit complicated.  
Instead, I would say: (assuming $V$ has local minimum at $X_1$): let $U_n$ be the connected component of $X_1$ in the open set $\{X: V(X)<V(X_1)+1/n\}$. The sets $U_n$ are invariant both ways, since $V$ is conserved. They form a basis of neighborhoods for $X_1$. 
