Classify all critical points of Hamiltonian system 
Consider the following system describing pendulum
$$\begin{align} & \frac{dx}{dt} = y, \\ & \frac{dy}{dt} = − \sin x. \end{align}$$
I need to classify all critical points of the system.

All critical points are of the form $(k\pi,0)$ for any $k \in \mathbb{Z}$
I know the Hamiltonian of the system $\displaystyle H(x,y)= \frac{y^2}2 -\cos x $, but I'm not sure if this can help in any way.
 A: For stability of stationary points you only need to look at the Jacobian matrix
$\begin{pmatrix} f_{x} && f_{y}\\
g_{x} && g_{y}
\end{pmatrix}$, where $f = y$ and $g = -sin(x)$.
For two dimensional dynamical system the behavior of stationary points are well-studied. This is summarized in the following diagram (from http://www.math24.net/equilibrium-points-of-linear-autonomous-systems.html):

Therefore you only need to examine the trace and determinant of the Jacobian matrix. 
Let's look at the stationary points $(k\pi, 0)$. 
For odd $k$ the Jacobian matrix is
$\begin{pmatrix} 0 && 1\\
1 && 0
\end{pmatrix}$. The trace is 0, and determinant is -1, so it is a saddle point.
For even $k$ the Jacobian matrix is
$\begin{pmatrix} 0 && 1\\
-1 && 0
\end{pmatrix}$. The trace is 0, and determinant is 1, so it is a center.
This can be verified by plotting the phase plane

A: Consider a small perturbation about the critical points $(x,y) = (0,n\pi)$ for $n\in\mathbb{Z}$, i.e. we take
$$x = n\pi + \delta x,~~~~y = \delta y$$
Then the dynamical system becomes, to first order in the perturbations,
$$\dot{\delta x} = \delta y~~~~\text{and}~~~~\dot{\delta y} = (-1)^{n+1}\delta x$$
which also implies $\ddot{\delta x} = \dot{\delta y} = (-1)^{n+1}\delta x$. If $n$ is odd then $\ddot{\delta x} = \delta x \implies \delta x \propto e^{\pm t}$ so perturbations grow exponentially and the fixpoint is unstable. If $n$ is even then $\ddot{\delta x} = -\delta x \implies \delta x \propto \sin(t+\phi)$ and $\delta y \propto \cos(t + \phi)$ so the size of the perturbations do not grow in time.
If you prefer a more physical approach: the system describes a particle rolling without friction in a potential $V(x) = -\cos(x)$ (since $\ddot{x} = -\frac{dV(x)}{dx}$), see figure below for a plot of the potential $V(x)$.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
When $n$ is odd we are at a peak of the potential so a slight perturbation will make the particle roll away from the fixpoint. When $n$ is even we are at the bottom of the potential so a slight perturbation away from the fixpoint will just make the particle oscillate around it (and since there is no friction the oscillations will not die out).
