Deriving the Poisson Integral Formula from the Cauchy Integral Formula

If $0 < |z| < 1$, $z$ lies inside of $\gamma$ and $1/\bar{z}$ lies outside of $\gamma$. So

$$\int_\gamma \frac{f(w)}{w - z}\, dw = 2\pi i f(z)$$

by Cauchy's integral formula and

$$\int_\gamma \frac{f(w)}{w - 1/\bar{z}}\, dw = 0$$

by Cauchy's integral theorem. This yields your first equation above.

Now

\begin{align}2\pi i f(z) &=\int_\gamma \frac{f(w)}{w - z} \, dw - \int_\gamma \frac{f(w)}{w - 1/\bar{z}}\, dw\\ &= \int_\gamma \left(\frac{1}{w - z} - \frac{1}{w - 1/\bar{z}}\right)f(w)\, dw\\ &=\int_\gamma \frac{z - 1/\bar{z}}{w^2 - (z + 1/\bar{z})w + z/\bar{z}}f(w)\, dw\\ &= \int_\gamma \frac{|z|^2 - 1}{\bar{z}w - (|z|^2 + 1) + zw^{-1}}f(w)\frac{dw}{w}\\ &= \int_\gamma \frac{|z|^2 - 1}{\bar{z}w + z\bar{w} - (|z|^2 + 1)}f(w)\, \frac{dw}{w}\\ &= \int_\gamma \frac{1 - |z|^2}{1 - 2\operatorname{Re}(z\bar{w}) + |z|^2}f(w)\, \frac{dw}{w}.\tag{1}\\ \end{align}

Letting $z = re^{i\theta}$ and parametrizing $\gamma$ by $w = e^{it}$, $0 \le t \le 2\pi$, we write

\begin{align} \int_\gamma \frac{1 - |z|^2}{1 - 2\operatorname{Re}(z\bar{w}) + |z|^2}f(w)\, \frac{dw}{w}&= \int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos(\theta - t) + r^2}f(e^{it}) i\, dt.\tag{2} \end{align}

By $(1)$ and $(2)$,

$$2\pi i f(re^{i\theta}) = \int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos(\theta - t) + r^2}f(e^{it}) i\, dt.$$

This is equivalent to

$$f(re^{i\theta}) = \frac{1}{2 \pi}\int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos(\theta - t) + r^2}f(e^{it}) \, dt.$$ $$$$