Proving critical point of a system is a center We have
$$ \begin{cases} \dot{x} = - y - y^3 \\ \dot{y} = x \end{cases} $$
where $x,y \in \mathbb{R}$. Show that the critical point for the linear system is a $\mathbf{center}$. Prove that the type of the critical point is the same for the $\mathbf{nonlinear}$ system.
TRY:
Notice $( \dot{x}, \dot{y} ) = (0,0)$ iff $x = 0 $, $y + y^3 =0 \iff y(1+y^2) = 0$. Thus, the only critical point is $(0,0)$. Lets linearize the system. The jacobian is
$$ J(x,y) = \left( \begin{matrix} 0 & -1 - 3y^2 \\ 1 & 0 \end{matrix} \right) $$
We have
$$ J(0,0) = \left( \begin{matrix} 0 & -1  \\ 1 & 0 \end{matrix} \right) $$
And the eigenvalues of this matrix are $\lambda = \pm i $ showing that indeed the critical point is a $\mathbf{center}$. But, Im stuck on showing that the same is true for the nonlinear system. How do we show the existence of closed orbits near the critical point?

The differential equation is the same as y"+(y+y^3)=0, which describes free vibration of an oscillator with a nonlinear spring. The system is conserve in energy. So its orbit is a closed curve.
