Here is what I have so far,
since f is analytic at z we know it is infinitely differentiable at z, so $f'(z) \not = 0$ and so we know that $Res(f'/f,z) = \frac{f'(z)}{f'(z)} = 1$ and any pole of higher order will be a multiple of this result so it will still be an integer. Is there a mistake in this reasoning, it seems a bit too simple but I think I covered all the possibilities for singularities of f'/f
The statement $Res\left(\frac{f'}{f}, z\right) = = \frac{f'}{f'}$ isn't quite correct. In general, if $g$ has an order $k$ pole at $z_0$ and $h$ has an order $k+1$ pole at $z_0$, then \begin{align*} Res\left(\frac{g}{h},z_0\right) &= (k+1)\frac{g^{(k)}(z_0)}{h^{(k+1)}(z_0)} \end{align*} If $f$ has an order $k+1$ pole at $z_0$, then $f'$ has an order $k$ pole at $z_0$. Therefore \begin{align*} Res\left(\frac{f'}{f},z_0\right) = (k+1)\frac{(f')^{(k)}(z_0)}{f^{(k+1)}(z_0)} = (k+1)\frac{f^{(k+1)}(z_0)}{f^{(k+1)}(z_0)}= k+1 \end{align*}