Am i correct in the way I am deciding whether or not subsets of $\Bbb R^3$ are surfaces or not? I just want to make sure that there isn't any gaps in my reasoning ( or flat out mistakes!) before I try to learn anymore about classifying subsets of $\Bbb R^3$ as surfaces, so to that end .....
Consider the subsets of $\Bbb R^3$
i) $S^2=\{(x,y,z)\in \Bbb R ^3 | x^2+y^2+z^2=1\}$
ii) $A=\{(x,y,z)\in \Bbb R^3|x^2+z^2=y^2\}$ 
In deciding whether or not these are surfaces I know we can use a corollary of the implicit function from calculus. 
Theorem :suppose $f:\Bbb R^3 \rightarrow \Bbb R$ is smooth and c is a regular value of $f$, with $f^{-1}(c)\neq(0,0,0)$, then $f^{-1}(c)$ is a surface in $\Bbb R^3$.
Here is my attempt at using the corollary:
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i)$S^2=\{(x,y,z) \in \Bbb R^3 |x^2+y^2+z^2=1\}$
let $f(x,y,z)=x^2+y^2+z^2$, clearly f is smooth and maps $\Bbb R^3 \rightarrow \Bbb R$.
$Df=(\partial f/\partial x, \partial f/\partial y,\partial f/\partial z)=(2x,2y,2z)$
so every point except the origin is a regular point. 
$\Rightarrow f(0,0,0)=0$ is the only critical value of f and so $f^{-1}(1)=x^2+y^2+z^2$
describes a surface as 1 is a regular value , and so $S^2$ is a surface.
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ii)let $f(x)=x^2+z^2-y^2$
again this is smooth with the proper mapping needed.
$Df=(2x,2z,-2y)$
so again the only critical point is the origin
$\Rightarrow f(0,0,0)=0$ is the only critical value 
so $f^{-1}(0)=x^2+z^2-y^2$ does not describe a surface so A is not a surface.

An intuitive argument as to why $S^{2}$ is a surface can be made as such: it can be covered with 6 co-ordinate patches (up,down,right,left,front & back), and thus each point on it lies in the image of one of the 6 coordinate patches.
As for the second part, your main error is using the implicit function theorem as an If and only if statement.
