Using Cauchy Integral Formula: $\small\displaystyle \int_c \frac{e^{-z^2}}{z^2}dz$

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I am going over the solutions to previous problems in order to prepare for a test. I am having a hard time understanding even basic applications of Cauchy's Integral Formula.

For example, I have tried to solve for the solution to the integral $$\int_c \frac{e^{-z^2}}{z^2}dz$$ where C is any positively-oriented simple closed contour surrounding the origin.

The solutions simply states that $$\frac{1}{2\pi i}\int_C \frac{e^{-z^2}}{z^2}dz = (e^{-z^2})'|_{z=0} = 0$$ by Cauchy.

How is Cauchy applied here? What are the steps taken to get from one side of the = to the other?

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Hint: By Cauchy Integral Formula we have

$$f'(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^2} dz$$

Let $z_0 = 0$ and $f(z) = e^{-z^2}$.