I understand that $ \begin{align} \oint_C (z-z_0)^n dz = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases} \end{align}$
And I understand that $ f^{(0)}(z)=\dfrac{1}{2\pi i}\oint_\gamma f(\xi)\frac{1}{\xi-z}d\xi$
Differentiating we get $ f'(z)=\dfrac{1}{2\pi i}\oint_\gamma f(\xi)\frac{1}{(\xi-z)^2}d\xi $
Clearly $n \neq -1$ So why is that last integral not zero?
On
The last integral is not zero because there is an "$f$" in the integrand, which changes the integral substantially. For instance, if you throw away the $f$, you get $$\frac{1}{2\pi i}\int_\gamma \frac{1}{(\xi - z)^2}d\xi,$$ but we see that this is the integral with $f(\xi) = 1$ identically: $$0 = \frac{d}{dz}1 = \frac{1}{2\pi i}\int_\gamma 1\frac{1}{(\xi - z)^2}d\xi.$$
In pariticular, if $\gamma$ is contour of the circle, with counter-clockwise orientation and $f(z) = z$, then $$1 = f'(0) = \frac{1}{2\pi i }\int_{\gamma}f(\xi)\frac{1}{(\xi - 0)^2}d\xi = \frac{1}{2\pi i }\int_{\gamma}\frac{1}{\xi}d\xi \neq 0.$$
Because of the presence of $f(\xi)$ there. You can't jump from$$\oint_C(\xi-z)^{-2}\,\mathrm d\xi=0\text{ to }\oint_Cf(\xi)(\xi-z)^{-2}\,\mathrm d\xi=0.$$