The answer is negative.<!--
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More precisely, $f:\RR\to\RR^n$ given by
$$ f(t) = \bigl( t,\abs{t}^{3+\frac{1}{2}},\abs{t}^{3+\frac{1}{3}},\ldots,\abs{t}^{3+\frac{1}{n}} \bigr) $$
gives an embedded $C^2$ curve in $\RR^n$ such that no neighbourhood of zero in the curve is contained in a smooth (or even $C^4$) hypersurface in $\RR^n$. By taking products of $f$ with any $\RR^l$ ($l\geq 0$), we produce examples of embedded $C^2$ submanifolds of $\RR^{n+l}$ of dimension $l+1$ for which there exists no neighbourhood of zero in the submanifold which is contained in a smooth hypersurface in $\RR^n$.

The argument below divides into two parts: the first involves only the local description of submanifolds of $\RR^n$; the second part involves an estimate using Taylor expansions. the proofs are elementary and will be given with some detail.


**Smoothly independent functions**


**Definitions:**

1. Let $f:\RR\to\RR$, $g:\RR\to\RR^n$ be functions $\RR\to\RR$. Say that $f$ is *depends smoothly* on $g$ at $t\in\RR$ if there exists a $C^\infty$ function $h:\RR^n\to\RR$ such that $f = h\circ g$ on some neighbourhood of $t$.
2. Call a function $f:\RR\to\RR^n$ *smoothly dependent* at $t\in\RR$ if, for some $1\leq i\leq n$, $f_i$ depends smoothly on $(f_1,\ldots,f_{i-1},f_{i+1},\ldots,f_n)$ at $t$. Otherwise, say $f$ is *smoothly independent* at $t$.

Let $f:\RR\to\RR^n$ be continuous. Let $L$ denote the image of $f$ as a subspace of $\RR^n$.


**Claim 1:**
Let $S$ be a $C^\infty$ submanifold of $\RR^n$ of codimension one (i.e. a smooth hypersurface) which contains some neighbourhood of $f(0)$ in $L$. Then $f$ is smoothly dependent at zero.

*Proof*:
<!--
Let $\pi_i:\RR^n\to\RR^{n-1}$ be the projection which forgets the $i$-th coordinate. Since $S$ is a hypersurface in $\RR^n$, there must exist $i\in\set{1,\ldots,n}$ such that $f(0)\in S$ is a regular point for $\pi_i|_S$. By the inverse function theorem, some neighbourhood $U$ of $f(0)$ in $S$ is parametrized as
-->
Since $S$ is a smooth hypersurface in $\RR^n$, it is locally the graph of a smooth function. More precisely, there exists $i\in\set{1,\ldots,n}$ such that some neighbourhood $U$ of $f(0)$ in $S$ is parametrized as
$$
U = \set{ (x_1,\ldots,x_{i-1},g(x),x_i,\ldots,x_{n-1}) \mid x \in V }
$$
where $V$ is an open in $\RR^{n-1}$ and $g:\RR^{n-1}\to\RR$ is a smooth function. Since $S$ contains some neighbourhood of $f(0)$ in $L$, we conclude that
$$
f_i(t) = g\bigl(f_1(t),\ldots,f_{i-1}(t),f_{i+1}(t),\ldots,f_{n-1}(t)\bigr)
$$
for $t$ sufficiently close to zero. &#9632;



**Examples of smoothly independent functions**

In view of the previous proposition, it remains to construct, for each $n>0$, a $C^2$ embedding $f:\RR\to\RR^{n+1}$ which is smoothly independent at zero.

**Claim 2:**
Take any finite sequence $1=a_0,a_1,\ldots,a_n$ of positive real numbers such that none of them can be written as a linear combination of the others with non-negative integer coefficients. Then the function
$$
f(t) = (t,\abs{t}^{a_1},\ldots,\abs{t}^{a_n})
$$
is smoothly independent at zero.

Here is a fairly explicit instance of claim 2 which recovers the example given at the beginning of the answer.

**Example:**
Pick some positive integer $l$ and a sequence of pairwise distinct real numbers $b_1,\ldots,b_n \in (0,1)$. Take $a_i = l+b_i$ for $i>0$ ($a_0 = 1$). Then the condition in claim 2 above is verified, hence the corresponding $f$ is smoothly independent. Moreover, this choice of $f$ is a $C^{l-1}$ embedding.

Claim 2 is a consequence of the next lemma.

**Lemma:** Let $a_0,\ldots,a_n$ be positive real numbers such that $a_0$ is not a linear combination of $a_1,\ldots,a_n$ with non-negative integer coefficients. There exists no smooth function $h:\RR^n \to \RR$ such that $t^{a_0} = h(t^{a_1},\ldots,t^{a_n})$ for all $t>0$ in some neighbourhood of zero.

*Proof*:
Consider the Taylor polynomial for $h$ of a sufficiently large degree $k$ together with the corresponding [Peano remainder term](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor.27s_theorem_for_multivariate_functions) (it actually suffices to take $k = \ceiling{ \frac{a_0}{ \min\set{a_1,a_2,\ldots,a_n} } }$). Since $f$ is $C^k$<!--(actually, it suffices that $f$ is $k$ times differentiable at $0$)-->, we have (in little-o notation)
$$
f(x) = \sum_{\abs{I}\leq k} a_I x^I + o\left(\sum_{\abs{I} = k} \abs{x^I}\right)
$$
where $I=(i_1,\ldots,i_n)$ ranges over $n$-tuples of non-negative integers, $\abs{I}=i_1+\ldots+i_n$, and $x^I = (x_1)^{i_1} \cdots (x_n)^{i_n}$. The little-o symbol above is considered as $x\to 0$. Replacing into $t^{a_0} = h(t^{a_1},\ldots,t^{a_n})$:
$$
t^{a_0} = \sum_{\abs{I}\leq k} a_I t^{i_1 a_1 + \cdots + i_n a_n} + o(t^{a_0})
$$
Given that $a_0$ is not a linear combination of $a_1,\ldots,a_n$ with non-negative integer coefficients, all the terms $a_I t^{i_1 a_1 + \cdots + i_n a_n}$ appearing above have degree different from $a_0$. Therefore, after rearranging and moving the monomials of degree greater than $a_0$ into the little-o symbol, we get
$$
t^{a_0} = \sum_{0\leq r<a_0} C_r t^r + o(t^{a_0})
$$
Here $r$ varies over the real numbers, but only finitely many coefficients $C_r$ are non-zero. In fact, all the coefficients $C_r$ are zero: if the lowest degree non-zero term on the right hand-side is $C_s t^s$ (with $0\leq s < a_0$), then
$$
C_s t^s = t^{a_0} - \sum_{s<r<a_0} C_r t^r - o(t^{a_0}) = o(t^s)
$$
and hence $C_s=0$. In conclusion, all the terms $C_r t^r$ are zero, and thus $t^{a_0} = o(t^{a_0})$, which is impossible. &#9632;