My answer is different from the one chegg.
Let C be the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ traversed once in the positive direction, and define
$$ G(z) :=\int_C \frac{\zeta^2-\zeta+2}{\zeta-z}\, d\zeta \hspace{8mm} \text{(z inside ${c}$)} $$ Find $G''(-i)$.
The following is my answer. $$ G''(-i) = \int_C \frac{\zeta^2-\zeta+2}{(\zeta + i)^3})\, d\zeta $$
$$ \begin{align} f(\zeta) &= \zeta^2-\zeta+2\\ f'(\zeta) &= 2\zeta-1\\ f''(\zeta) &= 2 \end{align} $$
$$ \begin{align} G''(-i) &= \frac{2 \pi}{2!} \cdot f''(-i) \\ &= \pi i \cdot 2 \\ &= 2 \pi i \end{align} $$
However, the correct answer from Chegg is $4\pi i$. where did I get wrong?
Since $f(z)=2z^2-z+1$, $f''(z)=4$ rather than $2$.
However, that's not where the problem lies. You should have got that$$G''(z)=2\oint_C\frac{\zeta^2-\zeta+2}{(\zeta-z)^2}\,\mathrm d\zeta.$$
On the other hand, note that$$G(z)=2\pi i(z^2-z+2),$$by Cauchy's integral formula. It's easier to solve the problem using this.