Answer
See below.
Work Step by Step
Let $z_1=a+bi$ and $z_2=c+di$. Then $\overline{\frac{z_1}{z_2}}=\overline{\frac{(a+bi)}{(c+di)}}=\overline{\frac{(a+bi)(c-di)}{(c+di)(c-di)}}=\overline{\frac{ac-adi+bci+bd}{c^2+d^2}}=\frac{ac+bd+(ad-bc)i}{c^2+d^2}$
$\frac{\overline{z_1}}{\overline{z_2}}=\frac{\overline{(a+bi)}}{\overline{(c+di)}}=\frac{a-bi}{c-di}=\frac{(a-bi)(c+di)}{(c-di)(c+di)}=\frac{ac+adi-bci+bd}{c^2+d^2}$. Thus they are equal, thus we proved what we wanted to.