Do superellipses provide examples of submanifolds of $\Bbb{R}^2$ that are not smooth?
Solution 1:
When $n\leqslant 1$, $f$ is not even a differentiable function. It is true that $f$ is smooth outside the set $(\mathbf{R}\times\{0\})\cup(\{0\}\times\mathbf{R})$, but $f^{-1}(1)$ crosses both of these lines, so it makes no sense saying that $1$ is a regular value of $f$.
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The preimage theorem you are talking about admits $C^k$ versions for all $k\in\mathbf{N}\cup\{\infty\}$, so your reasoning is true when $f$ is a $C^1$-map. The proof of the preimage theorem only requires the implicit function theorem to work, whose proof basically boils down to a Newton's method. So there is nothing specific about smooth maps to it.
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A critical level set of a function can be a smooth submanifold, but in this case the equation is said to be non-regular and the manifold is said to be non-transversally cut out. The easiest counterexample would be $f^{-1}(0)$ for $f(x,y)=x^2$.
It should be clear from the pictures drawn on Wikipedia that when $n<1$, the corresponding superellipse is not a smooth (nor $C^1$) submanifold of $\mathbf{R}^2$. However, your line of reasoning applies to the $n>1$ case.
Aside remarks.
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This a theorem by Kervaire and Milnor that all topological manifolds of dimension less or equal than $3$ can be endowed with a smooth structure (they are counterexamples in higher dimensions). However, this theorem is not saying that a topological submanifold of $\mathbf{R}^n$ of dimension less or equal than $3$ is a smooth manifold for the smooth structure induced by the one of $\mathbf{R}^n$.
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I should also be saying that if a topological space admits a $C^1$-structure, then it admits a smooth structure. Again, it is not saying that it is a smooth manifold for the same differentiable structure.
Solution 2:
Take the case $n=3$.
$$
f(x) = \big(1 - |x|^3\big)^{1/3}
$$
This is the graph of $f'''(x)$
$f'''(x)$ has a jump discontinuity at $x=0$.