Is $\epsilon^2/\epsilon^2=1$ or $0/0$? Is it possible in the system of dual numbers ($a+\epsilon b$; $\epsilon^2=0$) to calculate $\epsilon/\epsilon =1$? How then does one deal with $\epsilon^2/\epsilon^2=1$ versus $\epsilon^2/\epsilon^2=0/0$?
The same question for infitesimal calculus using hyperreal numbers where: $\epsilon \neq 0$ but $\epsilon^2=0$?
I probably did not use the correct formulation w.r.t. hyperreal numbers. I meant the axiom (?) in smooth infinitesimal analysis where it is assumed: $\epsilon \neq 0$ but $\epsilon^2=0$.
I am not quite sure how this analysis is related to nonstandard-analysis and hypercomplex numbers. I came across this topic in the book: A Primer of infinitesimal analysis (John L. Bell).

In the dual numbers, ${\mathbb R}[\epsilon]$ ($={\mathbb R}[X]/(X^2)$), $\epsilon$ is not invertible, so the expression $\epsilon / \epsilon$ ($= \epsilon \epsilon^{-1})$ is undefined.
In hyperreals, as Asaf Karagila mentions in the comments, $\epsilon^2 \neq 0$. There you do have $\epsilon / \epsilon = \epsilon^2 / \epsilon^2 = 1$ (as the hyperreals are a field and $\epsilon$ is a non-zero element).
I had a very quick look at the book by Bell. That's not only using infinitesimals, but also a different kind of logic (no law of excluded middle!). That's not for the faint-of-heart :-): for a given $x$, the statement "$x = 0 \lor x \neq 0$" is not necessarily true in that setting.
