How can I prove that?
I know if I do $\frac{d}{dz}(f(z)/(z-z_0)) = \frac{f'(z)}{z-z_0} - \frac{f(z)}{(z-z_0)^2}$, but if I put this derivative as zero, I will have these two terms on the right being equal.
Can I then just put that term in the Cauchy integral and say that they are the same?
The question: http://www.math.ubc.ca/~cautis/calcIV02/cpx3.pdf -> page 12, ex 9b
You need to use the fact that C is a closed curve (which is marked by the symbol $\oint$.
For an open curve from $z_1$ to $z_2$ we'd have $$ \int_C \frac{d}{dz} \big(\frac{f(z)}{z-z_0}\big) dz = \frac{f(z_2)}{z_2-z_0} -\frac{f(z_1)}{z_1-z_0}$$ but for a closed curve we have $z_1=z_2$, so $$ \oint_C \frac{d}{dz} \big(\frac{f(z)}{z-z_0}\big) dz = 0$$ Calculating the derivative you get the equality you want.
Another way: If $f(z)$ is holomorphic in the whole region bounded by $C$ (which we didn't have to assume in the previous method) we can use the Cauchy formula $$ \oint_C \frac{g(z)}{z-z_0}dz = g(z_0)$$ with $g(z)=f'(z)$; then we get $$ \oint_C \frac{f'(z)}{z-z_0}dz = f'(z_0)$$ On the other hand $$ f'(z_0) = \frac{d}{dz_0} f(z_0) = \frac{d}{dz_0} \oint_C \frac{f(z)}{z-z_0}dz = \oint_C \frac{\partial}{\partial z_0}\frac{f(z)}{z-z_0} dz= \oint_C \frac{f(z)}{(z-z_0)^2} dz$$