What is a simple example of a reduced, noetherian, local ring of dimension $0$ which is not Gorenstein? As the title says, I am looking for a noetherian local ring $R$ of dimension 0 which is reduced (and thus Cohen-Macaulay) but not Gorenstein.
Due to Bruns, Herzog $-$ Cohen-Macaulay Rings Theorem 3.2.10 every noetherian local ring which is not Gorenstein fails to be Cohen-Macaulay or fails to be of type 1. Since every reduced ring of dimension $\leq 1$ is Cohen-Macaulay (see Stacks-Reference), we are thus looking for a noetherian, reduced local ring of dimension 0 that fails to be of type 1.

What constitutes a simple example of such a ring?

I am grateful for any kind of help or input! Cheers!
 A: If you mean Krull dimension $0$, then I guess there is no example.
A reduced ring with Krull dimension $0$ is von Neumann regular, and a  local VNR ring is a field.
The reducedness condition really kills things.  $F_2[x,y]/(x,y)^2$ satisfies everything you said except it is not reduced.
A: Trying to read between the lines a little bit, it looks to me like you're looking for simple examples of Cohen-Macaulay local rings which are not Gorenstein, equivalently, as you say, that do not have type 1. In particular, you are looking to rings of small dimension to help you meet the Cohen-Macaulay condition.
While it is true that reduced Noetherian local rings of dimension at most $1$ are Cohen-Macaulay, this is really only relevant in dimension 1, as every Notherian local ring of dimension $0$ is Cohen-Macaulay, and, as others have already mentioned, a Noetherian local ring of dimension $0$ is Artinian, and an Artinian reduced local ring is a field.
The example given by rschiwieb in their answer is exactly the sort of example you seek; if k is a field, the ring $R=k[x,y]/(x,y)^2$ has type $2$.
This example can be extended to a larger family of examples:
Let $k$ be a field and let $R=k[x_1,\dots,x_n]/(x_1,\dots,x_n)^m$ where $n,m \ge 2$. Then $R$ is a Noetherian local ring of dimension $0$ (thus is Cohen-Macaulay) and the type $r(R)$ of $R$ is not $1$. In fact, one can compute explicitly that $r(R)=\mu_R[(x_1,\dots,x_n)^{m-1}]={m+n-2\choose n-1}$.
