Let $f^n$ be a family of $C^1$ functions and $f(x)=-|x|^2+1$ such that
$$f^n\to f$$
in $C^1$ sense as $\varepsilon\to 0$. I want to ask that does the level set $\{f^n=0\}$ converges to $\{f=0\}$ in some sense as $\varepsilon\to 0$? Is $\{f^n=0\}$ still a $C^1$ curve for $\varepsilon $ sufficiently small?



Convergence of a level set when $f^n\to f$ in $C^1$ sense
$\newcommand\de\delta$Apparently, (i) by $\epsilon\to0$ you meant $n\to\infty$ and (ii) by "$f^n\to f$ in $C^1$ sense" you meant that
$$\sup\_x|f^n(x)-f(x)|+\sup\_x|\nabla f^n(x)-\nabla f(x)|\to0. \tag{1}$$
If so, then the answers to both of your questions are positive:



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> Question 1: "does the level set $\{f^n=0\}$ converges to $\{f=0\}$ in some sense as $\varepsilon\to 0$?"
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Answer 1: Take any $\de\in(0,1]$. If $|x|\le1-\de$, then $f(x)=1-|x|^2\ge1-(1-\de)^2\ge\de$. So, by (1), eventually (that is, for all large enough $n$) $f^n(x)>0$ for all $x$ with $|x|\le1-\de$. Similarly, if $|x|\ge1+\de$, then $f(x)=1-|x|^2\le1-(1+\de)^2\le-2\de$. So, eventually $f^n(x)<0$ for all $x$ with $|x|\ge1+\de$. So, eventually the set $\{f^n=0\}$ will be $\de$-close to the set $\{f=0\}=\{x\colon|x|=1\}$.



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> Question 2: "Is $\{f^n=0\}$ still a $C^1$ curve for $\varepsilon $ sufficiently small?"
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Answer 2: The gradient $\nabla f(x)=-2x$ is bounded away from $0$ for all $x$ in a small enough open neighborhood, say $U$, of the set $\{f=0\}$. So, by (1), eventually the gradient $\nabla f^n$ will be bounded away from $0$ in $U$. Moreover, by Answer 1, eventually the set $\{f^n=0\}$ will be contained in $U$. So, eventually the gradient $\nabla f^n$ will be bounded away from $0$ in an open neighborhood of the set $\{f^n=0\}$. So, eventually $\{f^n=0\}$ will be a $C^1$ curve, by the [implicit function theorem](https://en.wikipedia.org/wiki/Implicit_function_theorem)


