Fleshing out (and slightly correcting) the trick mentioned in the comments:
$z\in\gamma\Longrightarrow 9=|z|^2=z\overline z\Longrightarrow \overline z=\frac{9}{z}\Longrightarrow$
$\oint_\gamma\frac{\overline z}{z-a}dz=9\oint_\gamma \frac{dz}{z(z-a)}dz =:9 I$
(1) If $\,|a|>3\,$ , then using Cauchy's Theorem:
$9I=9\oint_\gamma\frac{\frac{1}{z-a}}{z}dz=9\cdot 2\pi i\left.\frac{1}{z-a}\right|_{z=0}=-\frac{18\pi i}{a}$
(2) If $\,|a|<3\,$ , then we can integrate over little circles $\,\gamma_0\,,\,\gamma_a\,$ around zero and $\,a\,$ resp. that do not pass through the other point and are completely contained within $\,\gamma\,$ , and get:
$9I=9\oint_{\gamma_0}\frac{\frac{1}{z-a}}{z}dz+\oint_{\gamma_a}\frac{\frac{1}{z}}{z-a}dz=18\pi i\left(\left.\frac{1}{z-a}\right|_{z=0}+\left.\frac{1}{z}\right|_{z=a}\right)=$
$=18\pi i\left(-\frac{1}{a}+\frac{1}{a}\right)=0$