Hartshorne gives the example of completing the local ring of the affine variety defined by $y^2 - x^2 - x^3$ in $\mathbb{A}^2$ at $(0,0)$ on page 35. The takeaway seems to be that when we complete the local ring at $(0,0)$, we get a ring that is isomorphic to the local ring of the reducible algebraic set $S$ defined by the union of $y = x$ and $y = - x$, corresponding to the fact that in a neighborhood of $(0,0)$, the nodal curve $y^2 - x^2 - x^3$ "looks like" $S$.
I'm confused because I don't see where the point $(0,0)$ is used in the argument. Of course we are completing the local ring at this particular point, but in verifying that the completion is what Hartshorne claims it is, I don't think I've used any special properties about $(0,0)$ on the curve $y^2 - x^2 - x^3$, which should be necessary since the completion is capturing local information.
My argument that the completion is isomorphic to $k[[x,y]] / (y^2 - x^2 - x^3)$, is essentially: let $\mathfrak{m}$ be a maximal ideal in a Noetherian ring $R$. Then by properties of completion, since $0 \to (f) \to R \to R / (f) \to 0$ is exact, so is $0 \to \hat{f} \to \hat{R} \to \widehat{R / (f)} \to 0$. Thus $\widehat{R / (f)} \cong \hat{R} / \hat{f}$. Now $\hat{f} = f \hat{R}$ by considering both as coherent sequences of elements in $R$, and seeing that the coherence conditions on each give the same set of sequences, so we ahve that $\widehat{R / (f)} \cong \hat{R} / f \hat{R}$, and identifying $f$ with its image in $\hat{R}$, $\widehat{R / (f)} \cong \hat{R} / (f)$. Then we have that $\hat{R} \cong \widehat{R_\mathfrak{m}}$ by the argument here. So taking $\mathfrak{m} = (x,y)$, $R = k[x,y]$, and $f = y^2 - x^2 - x^3$, we get the desired result. Nothing in this argument seems to use any particular facts about the point $(0,0)$ or its maximal ideal, so why don't we get the result in Hartshorne that the local ring is isomorphic to the local ring of $S$ at $(0,0)$ for any point $P$ on $y^2 - x^2 - x^3$?
The point $(0,0)$ is used when you compute what the completion $\widehat{R}/(f)$ is. If you wanted to look at $(-1,0)$, you would consider $k[[(x+1),y]]/(y^2-x^2-x^3)$, which is isomorphic to $k[[y]]$ as follows: $y^2=x^3+x^2$ gives that $(x+1)=\frac{y^2}{x^2}$, so $(y^2-x^2-x^3)=((x+1)-\frac{y^2}{x^2})$ as $x=(x+1)-1$ is a unit in $k[[(x+1),y]]$ since it has nonzero constant term.