= Solution
<Dulac theorem> states the following. Let $D$ be a simply connected region for the planar system $\dot z=f(z)$, and suppose there is a continuously differentiable function $B:D\to\mathbb R$ for which
$$
\nabla\mathbin{\cdot}(Bf)
$$
has one strict sign throughout $D$. Then there is no <periodic orbit> contained in $D$.
To prove it, suppose a periodic orbit $\Gamma$ existed and let $A$ be its interior. Simple connectedness ensures $A\subset D$. By the planar <divergence theorem>,
$$
\iint_A\nabla\mathbin{\cdot}(Bf)\,dA
=\int_\Gamma Bf\mathbin{\cdot}n\,ds.
$$
The vector field $f$ is tangent to its trajectory $\Gamma$, so its scalar product with the outward normal $n$ vanishes. The right-hand side is therefore zero. The left-hand side cannot be zero because its integrand has one strict sign, a contradiction.
The <Poincaré-Bendixson theorem> states that if a forward trajectory of a smooth planar system remains in a compact set and its nonempty <omega-limit set> contains no equilibrium point, then that omega-limit set is a periodic orbit.
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