What I want to prove is the following:
Given $R$ a commutative unit ring which is also an integral domain, $I \subset R[x_1, \dots, x_n]$ an ideal, $f \in A = R[x_1, \dots, x_n]/I$ then: $$A_f = R[x_1, \dots, x_n, x_{n+1}]/J$$ Where $J$ is ideal generated in $A$ by $I$ and $fx_{n+1} - 1$, meaning with $f$ a representative.
I divided the proof in two parts.
For every ring commutative ring $R$ which is also an integral domain and for every $f \in R$: $R_f = R[x]/(fx - 1)$. To prove this I prove that $R[x]/(fx-1)$ has the universal property of the localization respect to the map $j = \pi \circ i$, where $i: R \hookrightarrow R[x]$ is the inclusion and $\pi$ is the projection on the quotient. Given a ring homomorphism $F : R \rightarrow B$ such that $F(f)$ is invertible we set $G: R[x] \rightarrow B$ such that $G(a) = F(a)$ for all $a \in R$ and $G(x) = (F(f)^{-1})$. Then $G(fx-1) = 0$ and so $G$ defines a map from $R[x]/(fx-1)$ to $B$ such that $G \circ j = F$. $G$ is also the unique map that shares this property because if another existed, call it $H$, then $1 = H(1) = H(xf) = H(x)H(f)$.
For every ring $R$, ideal $I\subset R$ and ideal $J \subset (R/I)[x]$ we have: $$\frac{(R/I)[x]}{J} \cong \frac{R[x]}{S}$$ Where $S$ is the ideal generated by $I$ and $\pi^{-1}(J)$, $\pi : R[x] \rightarrow (R/I)[x]$ being the projection. Here the obvious map is an isomorphism.
Finally: $$A_f = A[x]/(xf - 1) = \frac{(R[x_1, \dots, x_n]/I)[x]}{(xf-1)} = R[x_1, \dots, x_{n+1}]/J$$
Is my proof correct? Thank you!