Prove: $\overline{(\frac{z_1}{z_2})}=\frac{\overline{z_1}}{\overline{z_2}}$ 
Prove: $$\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}$$ 

$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{z_1\cdot z_2^{-1}}=\overline{z_1}\cdot \overline{z_2^{-1}}=\overline{z_1}\cdot \overline{z_2}^{-1}=\frac{\overline{z_1}}{\overline{z_2}}$$
Just to be sure is it valid?

Let's do this one in two parts.

If $z$ and $w$ are complex numbers, then $\overline{zw}=\overline{z}\,\overline{w}$.

Suppose $z=a+bi$ and $w=c+di$ with $a,b,c,d$ real numbers. Then
$$
\overline{zw} = \overline{(a+bi)(c+di)} = \overline{ac-bd + i(ad+bc)}
= ac-bd-i(ad+bc)
= (a-bi)(c-di)
= \overline{z}\,\overline{w}.
$$

If $z$ is a nonzero complex number, then $1/\overline{z}=\overline{(1/z)}$.

Suppose $z=a+bi$ with $a$ and $b$ real numbers such that $(a,b)\ne(0,0)$. Then
$$
\overline{\left(\frac{1}{z}\right)}
= \overline{\left(\frac{1}{a+bi}\right)}
= \overline{\left(\frac{a-bi}{a^2+b^2}\right)}
= \overline{\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i}
= \frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}i
= \frac{a+bi}{a^2+b^2}
= \frac{1}{a-bi}
= \frac{1}{\overline{z}}.
$$

Now
$$
\overline{\left(\frac{z}{w}\right)}
= \overline{\left(z\frac{1}{w}\right)} = \overline{z}\overline{\left(\frac{1}{w}\right)}
= \overline{z} \frac{1}{\overline{w}}
= \frac{\overline{z}}{\overline{w}}.
$$
This is essentially what you did.
