Let me turn my comment into a partial answer. My reading of the question is about *sufficiently small* open balls $U \subseteq \mathbf C^n$, because the question should be of a local nature.

**Lemma.** *If $X$ is smooth of dimension $d$ at $x$, then $(U,X \cap U,x)$ is homeomorphic to $(\mathbf C^n,\mathbf C^d,0)$ for $U$ sufficiently small, where $\mathbf C^d \hookrightarrow \mathbf C^n$ is a linear embedding.*

*Proof.* Since $X$ is smooth of dimension $d$ at $x$, it is a complete intersection in a Zariski open neighbourhood $V$ of $x$. That is, there exists a polynomial map $f \colon V \to \mathbf C^{n-d}$ such that $f^{-1}(0) = X \cap V$. Since $\mathrm df$ has maximal (real) rank $2(n-d)$ at $x$, the same holds in a small neighbourhood $U$ of $x$, so $U \to \mathbf C^{n-d}$ is a submersion and the result follows from the rank theorem. $\square$

So the question is whether this is a sufficient condition. The cuspidal curve already shows that merely remembering $(X \cap U,x)$ is not always enough, but I suspect that the full triple might be enough. Here is a positive result (ruling out the counterexamples so far):

**Lemma.** *Let $X$ be a hypersurface such that $U \setminus X$ is homotopy equivalent to a circle (for $U$ sufficiently small). Then $X$ is smooth at $x$.*

The hypothesis is in particular satisfied if $(U, X \cap U, x) \cong (\mathbf C^n,\mathbf C^{n-1},x)$ for a linear embedding $\mathbf C^{n-1} \hookrightarrow \mathbf C^n$.

*Proof.* Suppose $X = f^{-1}(0)$ for some $f \colon \mathbf C^n \to \mathbf C$. If $U$ is a small ball around $x$, then $E = U \setminus Z$ has a locally trivial Milnor fibration $f \colon E \to \mathbf C^\times$ with *Milnor fibre* $F$. We get a long exact sequence of homotopy groups
$$\ldots \to 0 \to \pi_1(F) \to \pi_1(E) \to \pi_1(\mathbf C^\times) \to \pi_0(F) \to \pi_0(E) \to \pi_0(\mathbf C^\times)$$
and isomorphisms $\pi_i(F) \stackrel\sim\to \pi_i(E)$ for $i \geq 2$. By assumption, $E$ is a $K(\pi,1)$, so $E \to \mathbf C^\times$ is a weak homotopy equivalence since $\pi_0(F) = *$. We conclude that $F$ is contractible, which implies that $x$ is a smooth point by a result of A'Campo [ACa73, Thm. 3]. $\square$

The proof uses nearby cycles, which for maps to higher-dimensional bases is a little more complicated (e.g. if $X$ is cut out by some map $f \colon \mathbf C^n \to \mathbf C^{n-d}$). Modern technology does allow us to consider that situation, but there might also be a more direct argument.

**Example.** For the cuspidal curve, we get $(U,X \cap U,x) \cong (\mathbf C^2,\mathbf C,*)$, but the embedding $\mathbf C\setminus * \hookrightarrow \mathbf C^2\setminus *$ is a (thickened) trefoil knot. So it is not isomorphic to the triple above.

(It is more traditional to consider $\partial U$ and $X \cap \partial U$, but then you can only talk about the two punctured strata and not all three strata. For hypersurfaces, we only needed the deleted link $U \setminus X$; I don't know if this is enough in general.)

**References.**

[ACa73] N. A’Campo, *Le nombre de Lefschetz d’une monodromie*. Indag. Math. (N.S.) **76**.2 (1973), p. 113-118. ZBL0276.14004.

1more comment