Multivariate residues in simple cases I have been looking at residues of multivariate functions and found there are quite a few difficulties (see e.g. Multivariate Residue Theorem? or Multivariate/multidimensional residues). In the literature, this is discussed in the context of manifolds, 1-forms and currents. Unfortunately, I am not an expert on manifolds.
Question: Are there "simple rules" deriving from the general treatment that can be applied in more basic cases.
I am thinking of multivariate functions $f(x,y,z, ...)$ with simple poles at equal points $x=y,~ x=z, ...$, where I would like to evaluate the residue at multiple, coinciding points $x=y=z$ as consecutive residues $\text{Res}_{x=y} \text{Res}_{y=z} \cdots$ in a consistent way.
Example: Consider the function $f(x,y,z) = \frac{1}{(x-y)(x-z)}$ defined on $\mathbb{C}^3$. It has singularities on and the 1-dimensional subspaces $\{(x,y,z) | x=y \}$ and $\{ (x,y,z) | x=z\}$ which intersect at $x=y=z$. Computing the residue on the intersection can be done through consecutive application of residues:
$$
\text{Res}_{y=z} \text{Res}_{x=y} \frac{1}{(x-y)(x-z)} = \text{Res}_{y=z} \frac{1}{(y-z)} = 1. 
$$
However, exchanging the residues leads to a wrong result:
$$
\text{Res}_{x=y} \text{Res}_{y=z} \frac{1}{(x-y)(x-z)} = 0.
$$
Is there a procedure that tells me how to correctly take certain residues or at least relate different combinations of residues which give the same result? (For example $\text{Res}_{y=z} \text{Res}_{x=y}$ and $\text{Res}_{z=x} \text{Res}_{y=x}$ in the previous example)
Background: In quantum field theory, amplitudes (vacuum expectation values of time-ordered products of fields) are meromorphic functions in $\mathbb{C}^k$. Poles correspond to the temporary fusion of particles and higher-order poles at the intersection of more than two points (which I would like to evaluate as consecutive residues) appear when there are more complicated composite particles.
I appreciate any help or literature recommendation!
 A: If think I have an idea which works in some cases at least. The idea is to cancel all poles and consider residue as limits of this (now) analytic function. This only works for simple poles of course. Here is what I mean:

Assuming that $f(x,y,z)$ has simple poles at $x=z$ and $y=z$. It follows that $(x-z)(y-z) f(x,y,z)$ is analytic around $x=z$ and $y=z$.
Therefore, it is possible to take limits of this continous function
\begin{align}
&\lim_{x\to z} \lim_{y\to z} (x-z)(y-z) f(x,y,z) = \lim_{y\to z} \lim_{x\to z} (x-z)(y-z) f(x,y,z) \\
\Leftrightarrow& \lim_{x\to z} (x-z) \lim_{y\to z} (y-z) f(x,y,z) = \lim_{y\to z} (y-z) \lim_{x\to z} (x-z) f(x,y,z)
\end{align}
The limits commute of course, because for an analytic function it does not matter from which direction we approach the point $x=y=z$.
These limits yield the residues at $x=z$ and $y=z$, so it follows that:
\begin{align}
&\text{Res}_{x=z} \text{Res}_{y=z} = \lim_{x\to z}(x-z) \lim_{y\to z} (y-z) = \lim_{y\to z} (y-z) \lim_{x\to z}(x-z) = \text{Res}_{y=z} \text{Res}_{x=z}.
\end{align}
This means: If one can exchange the limits consistently, one can exchange the residues.

In the example I gave, the function $f(x,y,z) = \frac{1}{(x-y)(x-z)}$ has poles at $y=x$ and $z=x$. Looking at the residue $\text{Res}_{y=z} \text{Res}_{x=y}$ means to consider $(y-z) (x-z) f(x,y,z)$. This is not analytic because the pole at $x=y$ is not cancelled. We can approach $x=y=z$ only by going first to $x=z$ and then $y=x$, not the other way around.
But $\text{Res}_{z=x} \text{Res}_{y=x} = \text{Res}_{y=x} \text{Res}_{z=x}$ can be exchanged consistently here.
