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.