I'm following this lecture note in which the definition of smooth map is given as follows.
Let $U \subset \mathbb{R}^{n}$ be open. Then a map $f: U \rightarrow \mathbb{R}^{m}$ is called smooth if all partial derivatives of all orders exist and are continuous.
Let $X \subset \mathbb{R}^{n}$ be instead an arbitrary subset. Then we say a map $f: X \rightarrow \mathbb{R}^{m}$ is smooth if $\forall x \in X, \exists$ an open subset $U \subset \mathbb{R}^{n}$ containing $x$ and a smooth map $\tilde{f}: U \rightarrow \mathbb{R}^{m}$ such that $\tilde{f}_{\restriction U \cap X}= f_{\restriction U \cap X}$.
It seems to me that an isolated point $x$ of $X$ does not affect the smoothness of $f$. This is because we can take a neighborhood $U_x$ of $x$ such that $U_x \cap X = \{x\}$. Then we define a map $\tilde{f}:U_x \to \mathbb R^m, y \mapsto f(x)$. Clearly, $\tilde{f}$ is smooth and agrees with $f$ on $U_x \cap X$.
Could you confirm if my understanding is correct?