Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$ 
Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$.

I don't know how to deal with this non-linear differential equation. I tried to consider $\ddot{x}(t+2\pi)+3x(t+2\pi)+x^3(t+2\pi)=0$ but with no success...
I need to prove this in order to solve a problem of dependence on initial conditions.
Can you please help me?

Letting $y=\dot{x}$ and $g(x)=3x+x^3$ we obtained the following first order planar system
$$
\begin{pmatrix} \dot{y} \\
                \dot{x}
\end{pmatrix} = \begin{pmatrix} -3x - x^3\\ y
\end{pmatrix} = \begin{pmatrix} - g(x) \\ y
\end{pmatrix} \tag{1}\label{one}
$$
This is a Hamiltonian system 
$$
\begin{pmatrix} \dot{y}\\ \dot{x} \end{pmatrix} =
\begin{pmatrix} -\partial_xH(x,y)\\ \partial_yH(x,y) \end{pmatrix}
$$
where $H(x,y)=\frac32 x^2 +\frac{x^4}{4}+\frac{1}{2}y^2=G(x) +\frac{1}{2}y^2$, where $G(x)=\int^x_0g$. Thus any solution $\mathbf{x}(t)=(x(t),y(t))$ to $\eqref{one}$ satisfies 
$$
H(\mathbf{x}(t))=H(\mathbf{x}_0)
$$
Here is a plot of a few level curves of the first integral $H$.

Since $G(x)= \frac32 x^2 + \frac14 x^4$ has a minimum at $x=0$, all solutions near the critical point $(0,0)$ of $\eqref{one}$ are periodic (this comes from Poincaré-Bendixon's theorem) and since $G(x)$ is even, a periodic solution crosses the $x$-axis at points $(\pm b,0)$ and the period of the solution is given by
\begin{aligned}
T_b&=2\int^b_{-b}\frac{1}{\sqrt{2(G(b)-G(x))}}\,dx=\frac{4}{\sqrt{2}}\int^b_0\frac{1}{\sqrt{G(b)-G(x)}}\,dx\\
&=4\sqrt{2}\int^b_0\frac{1}{\sqrt{(b^2+3)^2-(x^2 + 3)^2}}\,dx
\end{aligned}
Since $\sqrt{(b^2+3)^2-(x^2+3)^2}=\sqrt{(b-x)(b+x)(6+b^2+x^2)}$, we have that
$$\frac{4\sqrt{2}}{\sqrt{6+2b^2}}\leq T(b)\leq \frac{8\sqrt{2}}{\sqrt{6+b^2}}<\frac{8}{\sqrt{3}}<2\pi$$
Here is a plot of the period the period of solutions to $\eqref{one}$ as a function of the intercept $b$ as well as per and lower bounds for its decay.

