Proving a commutative ring can be embedded in any quotient ring.

The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.

Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:

$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\ \rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\ & & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0 \end{eqnarray}$

Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.

For cultural background, for an outstanding introduction to universal ideas see George Bergman's An Invitation to General Algebra and Universal Constructions.

You might also find illuminating Paul Cohn's historical article [Localization in general rings, a historical survey][3] - as well as other papers in that [volume [1]][4].

[1] Ranicki, A.(ed). [Noncommutative localization in algebra and topology][4]. ICMS 2002 [4]:http://www.maths.ed.ac.uk/~aar/books/nlat.pdf [3]:http://www.maths.ed.ac.uk/~aar/books/nlat.pdf#page=15

I'd say you're essentially on the right track, but I think you should think more carefully about how to rigorously define "$a/b$" (take a look at the page on localization if you need help). The "quotient ring" the question is asking about is called the total ring of fractions of $\mathfrak{R}$, which is a field if and only if $\mathfrak{R}$ is a domain.

Side note: I prefer the terminology "ring of fractions" (or "field of fractions") to "quotient ring", which to me is a term that refers to a quotient of a ring by an ideal.