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).

"Multiplication by $\epsilon$" gives an isomorphism $\mathbf{R} \to \epsilon \mathbf{R}[\epsilon]$ (i.e. the ideal of multiples of $\epsilon$ in the dual numbers). Thus it's not terribly abusive to write "division by $\epsilon$" as the inverse of this isomorphism. If you do so, then $\epsilon / \epsilon$ is the real number 1 (and not the dual number $1$).
More generally, in any one-dimensional vector space over $\mathbf{R}$, it is not terrible to define division of vectors to be the unique real number that satisfies the equation $(v/w)w = v$.
Generalizing in another direction, if $I$ is a principal ideal of a ring $R$ generated by $t$, then "multiplication by $t$" gives a surjective $R$-module homomorphism $\mu_t : R \mapsto I : r \mapsto tr$ (this is not a ring homomorphism!). Correspondingly, it is not terrible to define "division by $t$" on $I$ to be the inverse isomorphism $I \mapsto R / \ker \mu_t$ that maps $tr \mapsto r$.
e.g. if I write $f(x) + O(x^n)$ to be the equivalence class of $f(x)$ modulo $(x^n)$ in the ring $\mathbf{R}[x]$, then we would have the appealing fact
$$ \frac{xf(x) + O(x^n)}{x} = f(x) + O(x^{n-1}) $$
