is $f$ analytic inside $C?$ 
If $$f(z_0)=\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_0}dz$$ for all point $z_0$ inside $C,$ is $f$ analytic inside $C?~(C:$ simple closed contour$)$

 A: More generally, if $C$ is any rectifiable curve (need not be closed) and $f$ is a continuous function on $C$ then $$z\mapsto\frac1{2\pi i}\int_C\frac{f(\zeta)}{\zeta-z}\,d\zeta$$ is analytic for $z\notin C$.
This can be proved in at least three ways: You can differentiate under the integral sign, or if $z_0\notin C$, write $$\frac1{\zeta-z}=\frac1{\zeta-z_0+z_0-z}=\sum_{n=0}^\infty\frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}}$$
and note that the sum converges uniformly for $z$ in a circle with center $z_0$ and radius less than the distance from $z_0$ to $C$.
The third way may be the most fun: Show that the integral of the function around a small closed path vanishes. You need nothing more fancy than the ability to switch the order of integration.
A: Any Curve Homologous to $0$, Then This Result Holds.
A: For simplicity, assume $C:|z-z_0|=r$ be a circle around $z_0$ and $|f(z)|\leqslant M$ for all $z\in C$. Then 
\begin{align*}
\Big|\frac{f(z_1)-f(z_0)}{z_1-z_0}\Big|
=&\Big|\frac{1}{z_1-z_0}\Big|\cdot\Big|\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_1}dz-\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_0}dz\Big|\\
=&\frac{1}{2\pi}\Big|\frac{1}{z_1-z_0}\cdot\int_Cf(z)\Big(\frac{1}{z-z_1}-\frac{1}{z-z_0}\Big)dz\Big|\\
\leqslant&\frac{1}{2\pi}\int_C|f(z)|\cdot\Big|\frac{1}{z_1-z_0}\Big|\cdot\Big|\frac{1}{z-z_0}-\frac{1}{z-z_1}\Big|\cdot|dz|\\
\leqslant&\frac{1}{2\pi}\int_C\frac{M}{|z_-z_0|\cdot|z-z_1|}\cdot|dz|\\
\leqslant&\frac{1}{2\pi}\cdot\frac{M}{r(r-\varepsilon)}\cdot L_C\rightarrow\frac{1}{2\pi}\cdot\frac{M}{r^2}\cdot2\pi r=\frac{M}{r}~~~(z_1\rightarrow z_0)
\end{align*}

Your statement is not true in general, the following is a counterexample: 
Let $C:z=e^{i\theta},~\theta\in[0,2\pi)$. Define $f(z)=\frac{1}{\theta}$ for $z\in C\setminus\{1\}$ and $f(1)=\frac{1}{2\pi}$. Consider the differentiation of the point $z_0=r\in(0,1)$. Let $z_1=r+\varepsilon$. We have 
\begin{align*}
&\Big|\frac{f(z_1)-f(z_0)}{z_1-z_0}\Big|=\frac{|f(z_1)-f(z_0)|}{\varepsilon}\\
=&\frac{1}{\varepsilon}\Big|\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_1}dz-\dfrac{1}{2\pi i}\int_C\dfrac{f(z)}{z-z_0}dz\Big|\\
=&\frac{1}{2\pi\varepsilon}\Big|\int_Cf(z)\Big(\frac{1}{z-z_1}-\frac{1}{z-z_0}\Big)dz\Big|\\
=&\frac{1}{2\pi\varepsilon}\Big|\int_0^{2\pi}\frac{1}{\theta}\cdot\frac{z_0-z_1}{(z-z_1)(z-z_0)}d(e^{i\theta})\Big|\\
=&\frac{1}{2\pi\varepsilon}\Big|\int_0^{2\pi}\frac{1}{\theta}\cdot\frac{(-\varepsilon)ie^{i\theta}}{(e^{i\theta}-r-\varepsilon)(e^{i\theta}-r)}d\theta\Big|\\
=&\frac{1}{2\pi}\Big|\int_0^{2\pi}\frac{1}{\theta}\cdot\frac{e^{i\theta}}{(e^{i\theta}-r-\varepsilon)(e^{i\theta}-r)}d\theta\Big|\rightarrow\frac{K}{2\pi}\int_0^{2\pi}\frac{d\theta}{\theta}\rightarrow\infty~~(\varepsilon\rightarrow0)
\end{align*}
