Exceptional Divisor of Node Blow-Up is Two Smooth Reduced Points This is Vakil 29.3 D, self-study.
We are to show that if $X$ is a variety over an algebraically closed field $k$ with a node at point $p$, that the blow-up of $X$ at $p$ gives a morphism
$$\beta: \tilde{X} \to X$$
such that the exceptional divisor $\beta^{-1}(p)$ consists of two smooth reduced points. I am not sure what $\tilde{X}$ is here, but I assume it is the proper transform.
We have defined a node to be a closed point $p \in X$ such that $\widehat{\mathcal O}_{X, p} \cong k[[x,y]]/(xy)$ as topological rings, where the completion is of course taken $\mathfrak m_{X, p}$-adically.
I don't really see how to parse this. The given hint says to use the fact that completion is flat, I assume to conclude that the natural map $\mathcal O_{X, p} \to \widehat{\mathcal O}_{X, p}$ is flat. I assume then that we are supposed to use this to conclude that the proper transform map
$$\beta: \tilde{X} \to X$$
is then flat. Then, we are supposed to use 24.2 P a), I think to conclude that
$$\operatorname{Bl}_pX \times_X \tilde{X} \cong \operatorname{Bl}_{\beta^{-1}(p)}\tilde{X}$$
and somehow combine this information to turn this into a calculation on $\widehat{\mathcal O}_{X, p} \cong k[[x,y]]/(xy)$. What calculation? And is everything I have said thus far correct?
 A: Let's get one thing out of the way at the start: $\tilde{X}$ is the blowup of $X$ at $p$, which is defined intrinsically as the relative proj of $\bigoplus\mathcal{I}_p^n$. If $X$ is embedded as a closed subvariety of some larger variety $Y$, then $\tilde{X}$ is exactly the proper transform of $X$ under the blowup of $Y$ at $p$ - we won't really use that during the reduction here, but it could be useful at the end depending on how you decide to attack the calculation with $k[[x,y]]/(xy)$ (the proof of this statement is probably somewhere in the chapter on blowups). Note that the blowup map is essentially never flat, so you're barking up the wrong tree there.

Now on to the main problem. Consider the sequence of maps $$\operatorname{Spec} \kappa(p) \to \operatorname{Spec} \widehat{\mathcal{O}_{X,p}}\to \operatorname{Spec} \mathcal{O}_{X,p} \to X.$$ We can base change by $\beta:\widetilde{X}=Bl_p X\to X$ to get the following diagram:
$$\require{AMScd}
\begin{CD}
\widetilde{X}\times_X \operatorname{Spec} \kappa(p)  @>>> \widetilde{X}\times_X\operatorname{Spec} \widehat{\mathcal{O}_{X,p}} @>>> \widetilde{X}\times_X\operatorname{Spec} \mathcal{O}_{X,p} @>>> \widetilde{X}\\
@VVV @VVV @VVV @VV{\beta}V\\
\operatorname{Spec} \kappa(p) @>>> \operatorname{Spec} \widehat{\mathcal{O}_{X,p}} @>>> \operatorname{Spec} \mathcal{O}_{X,p} @>>> X
\end{CD}
$$
Let's make some observations.
First, $\operatorname{Spec} \mathcal{O}_{X,p} \to X$ is flat, because it's the identity on stalks.
Next, $\mathcal{O}_{X,p}\to \widehat{\mathcal{O}_{X,p}}$ is flat because completion is flat, so $\operatorname{Spec} \widehat{\mathcal{O}_{X,p}} \to \operatorname{Spec} \mathcal{O}_{X,p}$ is flat and therefore $\operatorname{Spec} \widehat{\mathcal{O}_{X,p}} \to X$ is flat.
By exercise 24.2.P(a), this means that $$\widetilde{X}\times_X\operatorname{Spec} \widehat{\mathcal{O}_{X,p}}\cong  Bl_{p\times_X \operatorname{Spec}\widehat{\mathcal{O}_{X,p}}} \operatorname{Spec} \widehat{\mathcal{O}_{X,p}}.$$
Finally, the top left corner is the fiber over $p$ in each of the other entries in the top row, so the fiber of $\beta:\widetilde{X}\to X$ over $p$ can be computed by looking at the fiber of $Bl_{p\times_X \operatorname{Spec}\widehat{\mathcal{O}_{X,p}}} \operatorname{Spec} \widehat{\mathcal{O}_{X,p}}\to \operatorname{Spec} \widehat{\mathcal{O}_{X,p}}$ over the closed point.
From here, you have some options: you can calculate the blowup of $(x,y)\subset k[[x,y]]/(xy)$ as Proj of $(k[[x,y]]/(xy))[t,u]/(xt-yu)$ with $x,y$ in degree zero and $t,u$ in degree one, or you can pull a fast one. We can replace $X$ with $\operatorname{Spec} k[x,y]/(xy)$ while keeping $\widehat{\mathcal{O}_{X,p}}$ the same, so the fiber over the node in general is the same as the fiber over the node in $\operatorname{Spec} k[x,y]/(xy)$. Both of these calculations should be approachable for you - a generalized version of the second computation can be found here, and it's very similar to what you'd need to do .
