I've been trying to prove this exercise for a while, and I've finally got something that vaguely looks right. For reference, the question is:
Let $X$ and $Y$ be two varieties. Suppose there are points $P \in X$ and $Q \in Y$ such that the local rings $\mathcal{O}_{P, X}$ and $\mathcal{O}_{Q, Y}$ are isomorphic as $\Bbbk$-algebras. Then show that there are open sets $P \in U \subseteq X$ and $Q \in V \subseteq Y$ and an isomorphism of $U$ to $V$ that sends $P$ to $Q$.
My attempt is as follows:
We assume that $X$ and $Y$ are both affine varieties, as we are just working locally. Let $\varphi: \mathcal{O}_Q \xrightarrow{\sim} \mathcal{O}_P$ be the isomorphism of local rings. We thus have that $\varphi(\mathfrak{m}_Q \mathcal{O}_Q) = \mathfrak{m}_P \mathcal{O}_P$ by definition of an isomorphism of local rings.
Because the field of fractions of $\mathcal{O}_P$ is $\Bbbk(X)$ and the field of fractions of $\mathcal{O}_Q$ is $\Bbbk(Y)$, we can extend $\varphi$ to an isomorphism of the fields of fractions $\varphi^*: \Bbbk(Y) \xrightarrow{\sim} \Bbbk(X)$. By corollary 4.5, we have that $\varphi^*$ defines a birational map from $X$ to $Y$. In other words, there are open subsets $U \subset X$ and $V \subset Y$ such that $\varphi^*$ induces a morphism $\psi: U \to V$. We want to show that we can take $U$ to be an open set containing $P$, that $\psi(P) = Q$, and that restricted to $U$, $\psi$ is an isomorphism from $U$ to $\psi(U)$.
Firstly, we show that we can take $U$ to be an open set containing $P$. Going through the construction of $U$ in Theorem 4.4, it suffices to show that if $y_1, \dots, y_m$ are the generators of $A(Y)$ as a $\Bbbk$-algebra, then $\varphi^*(y_i)$ is defined at $P$ for all $i$. But $\varphi^*(y_i) = \varphi(y_i)$ as $y_i \in A(Y) \subset \mathcal{O}_Q$, and so $\varphi(y_i) \in \mathcal{O}_P$, and so they are indeed defined at $P$. Thus, we can take $U$ to be an open set containing $P$. Similarly, we can take $V$ to be an open set containing $Q$ by considering the inverse of $\varphi^*$.
Now we show that $\psi(P) = Q$. For ease of notation, here we take $P$ and $Q$ to be the origin in their respective affine varieties by translation. Recall how $\psi$ is defined in Proposition 3.5: Let $A(Y) = \Bbbk[y_1, \dots, y_m]/I(Y)$ and let $\overline{y_i}$ be the image of $y_i$ in $A(Y)$. Then, consider $\xi_i = \varphi^*(\overline{y_i}) = \varphi(\overline{y_i}) \in \mathcal{O}_P$. Then, $\psi$ is defined by $\psi(T) = (\xi_1(T), \dots, \xi_m(T))$ for $T \in U$. By definition, $y_i \in \mathfrak{m}_Q$, so by the local ring homomorphism property, $\xi_i = \varphi(y_i) \in \mathfrak{m}_P \mathcal{O}_P$. Thus, $\xi_i(P) = 0$ for all $i$ by definition of $\mathfrak{m}_P$, and so $\psi(P) = (0, \dots, 0) = Q$.
Finally, we have an isomorphism of varieties from $\psi^{-1}(\psi(U)) \to \psi(\psi^{-1}(V))$ as in corollary 4.5, so we will just let $U$ and $V$ be these two open sets. With all of the above, we see that $\psi$ is an isomorphism from $U$ to $V$, and moreover $\psi(P) = Q$ as required.
The reason I doubt it is the section where I show $\psi(P) = Q$. I had the idea to translate $P$ and $Q$ to the origin because other solutions online do this. However, I don't actually think I really used this fact, did I? I could have just had $Q = (1, 2, 3, ...)$ and considered $y_i - i$ instead, for example, and I don't think anything about my proof would change. I would still show that $\psi(P) = (0, \dots, 0)$ instead of $\psi(P) = Q$.
Maybe there are more flaws that I can't sense either, I still feel like my algebraic geometry is weak. I post this proof because many of the other ones on this site and seem incorrect or are very unintuitive (such as the main post here: Problem I.4.7 in Hartshorne). I would very much appreciate some guidance here.
Thanks so much!