As answered in:
Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$
by user Empy2 in comments, to prove that a differential equation has periodic solutions, you must:
Find an $f(x,y)$ for which $f(x,\dot{x})=c$ is constant
Convert that to $\dot{x}=g(x)$
The period is $t=\int \frac{dx}{g(x)}$
If this is true, how can it be proved?
Thanks
Just to get periodic solutions it is sufficient to find the energy function $H(x,\dot x)=\frac12\dot x^2+G(x)$ with $G(0)=0$, with a square root $g(x)$ of $2G(x)=g(x)^2$ that is strictly monotonous on some interval $[-q,q]$. Then select $C=\min(G(p),G(-p))$ and conclude that all level curves $H(x,v)=c < C$ are bounded, thus closed. As $(x,v)=(0,0)$ is the only stationary point, on any level curve the tangents are non-zero, so after finite time the full level curve gets traversed. This gives a periodic solution.
In formulas, in the current example one gets $g(x)=x\sqrt{3+\frac12x^2}$, which is strictly monotonous, unbounded and smooth.
On a level set $2H(x,v)=2c=r^2$ the points $(g(x),v)$ follow a circle of radius $r$. Using polar coordinates for the circle, $$ g(x)=r\cos(\phi), ~~ \dot x=v=r\sin(\phi) \\~\\ g'(x)\dot x=-r\sin(\phi)\dot\phi \\~\\ g'(x)=-\dot\phi $$ Thus $\dot\phi=-g'(g^{-1}(r\cos\phi))$ is always negative.