How to determine the local ring $\mathcal{O}_{X,p}$ of the variety $X=V(xy)\subset\mathbb{A}^2$ at $p=(0,0)$. As in the title I am trying to solve the following execrcise: 

Compute the local ring $\mathcal{O}_{X,p}$ of the variety $X=V(xy)\subset\mathbb{A}^2$ at $p=(0,0)$.
  Show it is isomorphic to $A=\{(f_1,f_2)|f_1(0)=f_2(0)\}\subset\mathcal{O}_1\oplus\mathcal{O}_2$ where $\mathcal{O}_i$ are two copies of the local ring of $\mathbb{A}^1$ at $p=0$, i.e. $\mathcal{O}_{\mathbb{A}^1,0}$

My attempt is to exploit the isomorphism, which holds for any affine variety$$K[X]_{\mathcal{m}_p}\cong \mathcal{O}_{X,p}$$ where $K[X]_{\mathcal{m}_p}$ is the localization of the algebra of regular functions on $X$, at the maximal ideal of the point $p$. 
Then since $I(X)=(xy)$ and $\mathcal{m}_p=(x,y)$ I get: 
$$\mathcal{O}_{X,p}\cong (\frac{K[x,y]}{(xy)})_{(x,y)}$$
Here I am getting a bit confused with quotients and localization so:
Doubt 1: Letting $z_1=x+(xy),z_2=y+(xy)$, is it true that $\mathcal{O}_{X,p}\cong (K[z_1,z_2])_{(z_1,z_2)}$ ? Something here does not convince me. 
Then I have $\mathcal{O}_i=\{\frac{f}{g}|f,g\in K[t], g(0)\neq 0\}$, but I am a bit clueless on how to find the desired isomoprhism. So:
Doubt 2: How to prove $\mathcal{O}_{X,p}\cong A$
Thanks in advance. 
 A: Let me expand my comments in to an answer.
First, localization is exact, and therefore commutes with quotients: if $0\to A\to B\to B/A\to 0$ is exact, then $0\to S^{-1}A\to S^{-1}B \to S^{-1}(B/A)\to 0$ is exact, and the final term is also isomorphic to $(S^{-1}B)/(S^{-1}A)$. In your case, this implies that $\left(K[x,y]/(xy)\right)_{(\overline{x},\overline{y})}\cong k[x,y]_{(x,y)}/(xy)_{(x,y)}$ where the overline denotes taking the image inside $k[x,y]/(xy)$. This should resolve your first question.
For your second question, one way to do it is to first understand the global picture: one can write $k[x,y]/(xy)$ as the subring of $k[x]\times k[y]$ of polynomials having equal constant term. Localization at the maximal ideal corresponding to the origin means inverting all polynomials on both sides which do not vanish at the origin: in particular, any polynomial in $k[x]\times k[y]$ of the form $(c+xp(x),c)$ or $(c,c+yq(y))$ is inverted, so we can write this localization as the subring of $k[x]_{(x)}\times k[y]_{(y)}$ consisting of polynomials with equal constant terms. As $k[x]_{(x)}$ is the local ring of $\Bbb A^1_k$ at the origin, we are done.
