It is well-known that the following Hamiltonian system 
\begin{eqnarray}
\left\{\begin{array}{rcl}
\frac{dx}{dt}&=&y,\\
\frac{dy}{dt}&=&x(-1+x^2),
\end{array}\right.
\end{eqnarray}
with 
$$ H(x,y)=\frac{x^2}2+\frac{y^2}{2}-\frac{x^4}{4}$$
has the solution
$$ x=\tanh\left(\frac{t}{\sqrt2}\right),\qquad y=\frac1{\sqrt2}\text{sech}^2\left(\frac{t}{\sqrt2}\right) $$
such that $H(x,y)=\frac14$. 


Now consider the following Hamiltonian system
\begin{eqnarray}
\left\{\begin{array}{rcl}
\frac{dx}{dt}&=&y,\\
\frac{dy}{dt}&=&x(-1+x^4),
\end{array}\right.\tag{1}
\end{eqnarray}
with 
$$ H(x,y)=\frac{x^2}2+\frac{y^2}{2}-\frac{x^6}{6}$$
and I want to find the explicit solution such that $H(x,y)=\frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated.



Explicit solution of a Hamiltonian system
Edited.


Surprisingly, there is an elementary solution, if I made no mistake in
the following computation.


Your equation is equivalent to
$$\left(\frac{dx}{dt}\right)^2=\frac{1}{3}(x^6-3x^2+2),$$
(I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.)
This equation is separable,
$$\frac{t}{\sqrt{3}}=\int\frac{dx}{\sqrt{x^6-3x^2+2}}=:I$$
and requires inversion of the integral.


To reduce it
to a standard integral, change $x^2=1/(u+1), \; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes
$$I=-\frac{1}{2}\int\frac{du}{u\sqrt{2u+3}}.$$
The inverse function to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution
$$x(t)=\frac{2\sinh(t+c)}{\sqrt{6+4\sinh^2(t+c)}}.$$
Can you check this computation?


Of course, your value $1/3$ for the Hamiltonian is crucial here. With
some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?


