For
$$
V(x,y)=\frac{x^2}{2}+g(y),
$$
the <orbital derivative> is
$$
\dot V=x(-2x+x^3+\sin2y)+g'(y)(-x-y^3).
$$
Choosing $g'(y)=\sin2y$ cancels the mixed term. Taking $g(0)=0$ gives
$$
g(y)=\int_0^y\sin(2s)\,ds
=\frac{1-\cos2y}{2}=\sin^2y,
$$
so
$$
\boxed{V(x,y)=\frac{x^2}{2}+\sin^2y}.
$$
This function has <positive definiteness of a Lyapunov function>[positive definiteness] near the origin, and
$$
\dot V
=-2x^2+x^4-y^3\sin2y
=-x^2(2-x^2)-y^3\sin2y.
$$
For $|x|<1$ and $|y|<1$, both terms are nonpositive and equality holds only at $(0,0)$. Thus $V$ is a <strict Lyapunov function> near the origin, and the <Second Lyapunov theorem> proves that the origin is asymptotically stable.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 31E / b / ii / Solution

A <Lyapunov function> is positive definite at an equilibrium $x_*$ when $V(x_*)=0$ and $V(x)>0$ for nearby $x\ne x_*$. Thus $x_*$ is a strict local minimum of $V$.

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 Incoming links: Positive definiteness of a Lyapunov function