Show that a cubic second order ODE only has periodic solutions

You found that along a solution the expression $$ \dot x^2+x^2(1+\tfrac12x^2)=R^2 $$ is constant. The left side can be seen a sum of two squares, so that the whole is a circle equation for radius $R$. Parametrize with the angle as $\dot x=R\cos(u)$, $x\sqrt{1+\frac12 x^2}=f(x)=R\sin(u)$. Then taking the derivative of the second expression and comparing with the first gives $$ f'(x(t))\dot x(t)=R\cos(u(t))\dot u(t)\\ \implies f'(x(t))=\dot u(t) $$ for any motion where $\dot x$ is not constant zero.

Now $f(x)$ is a strictly increasing function, $f'(x)\ge 1$ is always positive. This means that $u(t)$ is also strictly increasing, and the phase curve $(x(t),\dot x(t))$ moves along the image of a circular motion, thus is periodic.

$$ f'(x)=\sqrt{1+\tfrac12 x^2}+\frac{x^2}{2\sqrt{1+\tfrac12 x^2}}=\frac{1+x^2}{\sqrt{1+\tfrac12 x^2}}\ge 1 $$ and $y=f(x)$ can be solved observing $\operatorname{sgn}(y)=\operatorname{sgn}(x)$ as follows $$ y^2=x^2+\tfrac12x^4\\ 1+2y^2=(1+x^2)^2\\ x^2=\sqrt{1+2y^2}-1=\frac{2y^2}{1+\sqrt{1+2y^2}}\\ \implies x=\frac{\sqrt2 y}{\sqrt{1+\sqrt{1+2y^2}}} $$