converse cauchy integral's formula: $F$ is holomorphic if $F(w) = \frac{1}{2\pi i} \oint_{\gamma} \frac{F(z)}{z-w} d z$? Let's explain it more detail.
Suppose that $U\subseteq \mathbb{C}$ is an open set. Let $F\in C^0(U)$. Suppose that for every $\bar{D}(z,r) \subseteq U$ and $\gamma$ the curve surrounding this disc (with counterclockwise orientation) and all $w\in D(z,r)$ it holds that
\begin{align}
  F(w) = \frac{1}{2\pi i} \oint_{\gamma} \frac{F(\zeta)}{\zeta-w} d\zeta
\end{align}
Textbook says that $F$ is holomorphic.
How do I prove this from the definition of holomorphic. $i.e$, $\frac{\partial F}{\partial \bar{w}}=0$.

Your hypothesis allow you to take the derivative inside the integral sign:
$$ \partial_{\bar{w}} F(w) = \frac{1}{2\pi i} \oint_\gamma F(\zeta) \left(\partial_{\bar{w}} \frac{1}{\zeta-w}\right) \; d\zeta = 0 $$
