= Solution
On the two vertical sides of the square, the $x$ component of the <vector field> satisfies
$$
\dot x\big|_{x=1}=-1+\sin(2y)\leq0,
\qquad
\dot x\big|_{x=-1}=1+\sin(2y)\geq0.
$$
On the horizontal sides,
$$
\dot y\big|_{y=1}=-x-1\leq0,
\qquad
\dot y\big|_{y=-1}=1-x\geq0.
$$
Thus the vector field never points outward through the boundary, so the closed square is a <positively invariant set> and trajectories cannot leave $S$.
At a fixed point, $\dot y=0$ gives $x=-y^3$. Substitution into $\dot x=0$ gives
$$
2y^3-y^9+\sin(2y)=0.
$$
For $0<y<1$, both $y^3(2-y^6)$ and $\sin(2y)$ are positive; for $-1<y<0$, both are negative. Hence $y=0$, and then $x=0$, is the unique <equilibrium point> in $S$.
This alone does not prove that $S$ lies in the <domain of stability>. A compact trapping region containing a unique equilibrium can also contain a <periodic orbit> or another nontrivial <omega-limit set>.
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