= Solution
Write $y=\dot x$ and $E=(x^2+y^2)/2$. Along a trajectory,
$$
\dot E=-\epsilon(\alpha x^2+\beta y^2-\gamma)y^2.
$$
Use the unperturbed orbit $x=R\sin t,y=R\cos t$ to compute the leading energy change over one period:
$$
\Delta E=-\epsilon\pi\left[\frac{\alpha+3\beta}{4}R^4-2\gamma R^2\right]+O(\epsilon^2).
$$
Zero averaged energy change gives the nonzero amplitude
$$
\boxed{R=2\sqrt{\frac{\gamma}{\alpha+3\beta}}+O(\epsilon).}
$$
The supplied sign condition makes this real and nonzero. The averaged <amplitude equation> is $\dot R=\epsilon[\gamma R/2-(\alpha+3\beta)R^3/8]+O(\epsilon^2)$, whose <derivative> at that root is $-\epsilon\gamma+O(\epsilon^2)$. Its nonzero <derivative> also makes the root persist as an isolated <periodic orbit> for sufficiently small positive $\epsilon$.
Directly, the planar <divergence> is $-\epsilon(\alpha x^2+3\beta y^2-\gamma)$. On the leading orbit its integral over $2\pi$ is $-\epsilon[\pi R^2(\alpha+3\beta)-2\pi\gamma]=-2\pi\epsilon\gamma$. Thus
$$
\boxed{\mu=\exp[-2\pi\epsilon\gamma+O(\epsilon^2)].}
$$
\b[The cycle is attracting for $\gamma>0$ and repelling for $\gamma<0$]. Both sign possibilities are allowed by $(\alpha+3\beta)\gamma>0$.
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