Can a Lyapunov Function's Derivative be Negative Semi-Definite Outside a Contour?

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A question I recently came across challenged the definition I learned, and keep finding, about Lyapunov Stability:

"Given the system: $$\frac{dx}{dt}=x(1-x^{2})+y$$ $$\frac{dy}{dt}=y(1-y^{2})-x$$ Prove the existence of a limit cycle. Hint: $v(x,y)=x^{2}+y^{2}$ may be helpful."

Using the fact we can assert the existence of a limit cycle with the Lyapunov Stability Theorem, I assumed $v(x,y)$ was the provided Lyapunov function, and went on to show $\frac{dv}{dt}$ is negative semi-definite $\forall (x,y)\neq(0,0)$, the fixed point (implying $(0,0)$ is stable and thus has a limit cycle rotating around). Showing the simplified results: $$\frac{dv}{dt}=2(x^{2}-x^{4}+y^{2}-y^{4})$$ But $\frac{dv}{dt}$ is only negative semi-definite outside the contour $x^{2}-x^{4}+y^{2}-y^{4}=0$.

Regarding the Lyapunov Stability Theorem, did the question statement give an incorrect Lyapunov function? Or am I misinterpretting the theorem altogether?

As always, a big thank you to those who spend their time helping to resolve and explain these questions.