= Solution
A form of the <Poincaré-Bendixson theorem> states that a nonempty <omega-limit set> of a forward orbit of a $C^1$ planar autonomous <vector> field, contained in a compact region and containing no <equilibrium>, is a <periodic orbit>.
Set $\rho=x^2+y^2$. Direct calculation gives
$$
\dot\rho=2\rho(1-\rho)-4k^2x^2y^2,
\qquad2\rho-(2+k^2)\rho^2\leq\dot\rho\leq2\rho(1-\rho),
$$
because $0\leq x^2y^2\leq\rho^2/4$. At the inner boundary $\rho=2/(2+k^2)$ the flow points into or along the annulus, and at the outer boundary $\rho=1$ it does likewise. Smoothness and these inequalities make the compact annulus forward invariant.
For nonzero $(x,y)$, write $x=\sqrt\rho\cos\theta$, $y=\sqrt\rho\sin\theta$. Then
$$
\dot\theta=\frac{x\dot y-y\dot x}{\rho}
=1-\frac{k^2\rho}{4}\sin4\theta\geq1-\frac{k^2}{4}>0
$$
throughout the annulus. Hence it contains no <equilibrium>. Any orbit starting there remains compactly trapped, has a nonempty <omega-limit set>, and the <Poincaré-Bendixson theorem> supplies the required \b[<periodic orbit> in the stated annulus]. For $k=0$ the annulus reduces to $\rho=1$ itself, and there $\dot\theta=1$, giving the <periodic orbit> directly.
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