Q:

$\mathcal{O}_{\mathbb{A}^n,x}\simeq \mathbb{C}[X_1,...,X_n]_{m(x)}$

I try to understand the proof of the following proposition:

For every point $x=(x_1,...,x_n)\in \mathbb{A}^n$ there exists a
  natural isomorphism 
$$\mathcal{O}_{\mathbb{A}^n,x}\simeq \mathbb{C}[X_1,...,X_n]_{m(x)}$$
where $m(x)=(X_1-x_1,...,X_n-x_n)$.

The proof starts with

There exists a natural map from $\mathbb{C}[X_1,...,X_n]_{m(x)}$ to
  $\mathcal{O}_{\mathbb{A}^n,x}$ how associates to every fraction the
  germ of  the function that it defined. 

An element from $ \mathbb{C}[X_1,...,X_n]_{m(x)}$  has the form $\frac{p}{q}$ where $p,q\in \mathbb{C}[X_1,...,X_n]$, $q(x)\neq 0$. 
On the other side $\mathcal{O}_{\mathbb{A}^n,x}=\bigcup\limits _{x\in U} {\lbrace f\in \mathcal{O}_{\mathbb{A}^n}(U)\rbrace}/\sim$, where $g\in \mathcal{O}_{\mathbb{A}^n}(U)\sim h\in \mathcal{O}_{\mathbb{A}^n}(V)$ iff there exists $W\subseteq U\cap V$ s.t.  $g=h$ on $W$.
The isomorphism can be written explicitly? 

A:

Let $X = \mathbb{A}^n$ and $\newcommand{\m}{\mathfrak{m}} \m = m(x)$.  Following the proof, given $f = \frac{p}{q} \in \mathbb{C}[X_1, \ldots, X_n]_\m$, then $f$ defines a regular function on the complement of the set of zeroes of $q$, which I'll denote $U_f$.  So $\DeclareMathOperator{\O}{\mathcal{O}} f \in \O_X(U_f)$, hence we can map it to its equivalence class in $\mathcal{O}_{X,x}=\bigcup_{x\in U} \{ f\in \mathcal{O}_{X}(U)\}/\sim$.  (Note that $x \in U_f$.)  This gives a map $\mathbb{C}[X_1, \ldots, X_n]_\m \to \O_{X,x}$.
Conversely, given an element of $\O_{X,x}$ we can choose a representative $f_1 \in \O_X(U)$ for it, where $U$ is some open set containing $x$.  Since $f_1$ is regular at $x$, then there exists some open set $U_x$ and polynomials $g_1, h_1 \in \mathbb{C}[X_1, \ldots, X_n]$ with $h_1(x) \neq 0$ such that $f_1 = g_1/h_1$ on $U_x$.  Since $h_1(x) \neq 0$, then $g_1/h_1 \in \mathbb{C}[X_1, \ldots, X_n]_\m$, so we have produced a map
\begin{align*}
\O_{X,x} &\to \mathbb{C}[X_1, \ldots, X_n]_\m\\
f_1 &\mapsto g_1/h_1 \, . 
\end{align*}
To show this map is well-defined, suppose $f_2$ is another representative, with $f_2 = g_2/h_2$ on some open set $V_x$.  Since $f_1 \sim f_2$, then there exists a nonempty open set $W_x \subseteq U_x \cap V_x$ such that $g_1/h_1 = f_1 = f_2 = g_2/h_2$ on $W_x$. Then $g_1 h_2 - g_2 h_1$ is a polynomial that is identically zero on the set $W_x$, and since $\mathbb{C}$ is infinite this implies that $g_1 h_2 - g_2 h_1$ is the zero polynomial.   Thus $g_1/h_1 = g_2/h_2$.
Now you just have to show that these two maps are mutually inverse.