Consider the following system
$ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} -x \\ (xy-1)y^3+(xy-1+x^2)y \end{bmatrix}$
The equilibrium point is $(0,0)$ and by linearizing the system we get
$J\vert_{(0,0)}=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$, $\rm tr(A)<0, \rm det(A)>0 \implies \lambda_1,\lambda_2<0$. Thus stable equilibrium point.
How can I show that it is not globally asymptotically stable? Moreover, we can see that $S=\{x,y\in\mathbb{R}^2: xy\geq 0\}$ is positively invariant.
Suppose $x(0) y(0) > 10$, and let $S = \{t \, : \, x(t) y(t) > 10\}$. By assumption $S$ is nonempty. If $T := \sup(S) < \infty$, then by continuity it must be that $x(T) y(T) \ge 10$.
We then compute $$(xy)'(T) = (xy-1)y^3 x + (xy-2 + x^2)yx \ge 90y^2 + 10(8 + x^2) > 0$$ so in fact there must exist $\epsilon > 0$ such that $x(T+\epsilon) y(T+\epsilon) > x(T) y(T) \ge 10$, contradicting the definition of $T$. Hence $T = \infty$, i.e. given that $x(0) y(0) > 10$, we have $x(t) y(t) > 10$ for all $t \ge 0$.
To conclude, note that since $x(t) = x(0)e^{-t}$ has $x(t) \to 0$ as $t\to\infty$, in order for $x(t) y(t) > 10$ to hold we must have $\limsup_{t\to\infty} y(t) = \infty$, so in particular any initial condition with $xy > 10$ cannot converge to $(0,0)$.