Consider $F(z)=\sum_i a_iz^i$ to be a formal power series with coefficients $a_i$. It is known that the coefficients of the series can be recovered from the $n$th terms of the associated Taylor series $$ a_n = \frac{1}{n!}\frac{d^nF}{dz^n}\bigg|_{z=0} $$ The computation of the $n$th derivative (for my purpose) proceeds via the Cauchy formula $$ \frac{d^nF}{dz^n} = \frac{n!}{2\pi i}\oint_\Delta \frac{F}{(z-a)^{n+1}}dz $$
However, how would this look in more than one dimension?
Consider now that $F(x,y)=\sum_{ij}a_{ij}x^iy^j$. The coefficient is given by
$$ a_{n,m} = \frac{1}{n!m!}\left[\frac{\partial }{\partial x}\right]^n\left[\frac{\partial }{\partial y}\right]^m F(x,y)\ \bigg|_{x=y=0} $$
So now, we have to identify the $n$ and $m$ th partial derivatives using the Cauchy formula.
$$ \left[\frac{\partial }{\partial x}\right]^n\left[\frac{\partial }{\partial y}\right]^m F(x,y)\ \bigg|_{x=y=0} $$
Can we nest the Cauchy formula as follows?
$$ \left[\frac{\partial }{\partial x}\right]^n\left[\frac{\partial }{\partial y}\right]^m F(x,y)\ \bigg|_{x=y=0} = \frac{n!m!}{(2\pi i)^2}\oint \frac{1}{(x-a_1)^{n+1}}\left(\oint \frac{F(x,y)}{(y-a_2)^{m+1}}dy\right)dx $$
Yes, you can fix for instance $y=y_0$ and see $F(x,y_0)$ as one complex variable function and you can apply the Cauchy formula for the $m$ derivatives in $x$ and you can do the same for $x$.
You can check the Gunning Rossi "Analytic functions of several complex variables" in case you need a reference.