= Solution
First assume $\omega_0\ne0$ and regard $\theta$ modulo $2\pi$. The section $\theta=0$ has return time $T=2\pi/|\omega_0|$. Solving the resonantly forced oscillator from this section gives
$$
x(t)=x_0\cos(\omega_0t)+\frac{y_0}{\omega_0}\sin(\omega_0t)+\frac{\gamma}{2\omega_0}t\sin(\omega_0t).
$$
At $t=T$, the sine vanishes and the cosine is one, while differentiating the secular term adds $\gamma T/2$ to $y$. Hence the <Poincaré map> is
$$
\boxed{P(x,y)=(x,y+\gamma T/2).}
$$
For $\gamma>0$, every iterate has $y_n=y_0+n\gamma T/2\to\infty$. Thus every oscillator trajectory is unbounded, regardless of its initial amplitude. For an arbitrary section phase $\theta_0$, the return map translates by $(\gamma T\sin\theta_0/(2\omega_0),\gamma T\cos\theta_0/2)$, which is still nonzero. At $\gamma=0$ the map is the identity, and all $(x,y)$ motions are bounded harmonic oscillations, including the zero-amplitude <equilibrium>.
\b[The zero-frequency case allowed by the printed parameter range is different.] If $\omega_0=0$, then $\theta=\theta_0$ is constant and $x=x_0+y_0t+\tfrac12\gamma\cos\theta_0t^2$. There is no phase-crossing Poincaré section. For $\gamma>0$, choosing $\cos\theta_0=0$ and $y_0=0$ gives bounded <equilibria>, contradicting an unqualified “all trajectories” statement. For $\gamma=0$ at zero frequency, free-particle motion is unbounded unless $y_0=0$. The <resonance> conclusion therefore requires the natural nonzero-frequency assumption.
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