Showing a function is a manifold

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I have just been introduced to the world of manifolds in my real analysis class, and I'm having some trouble really understanding what manifolds are and showing why they exist. I have been given the seemingly simple problem to show:

The surface $x^2 - y^2 = 0$ consists of two planes whose intersection is the z axis. If the z axis is removed then what is left is a smooth two dimensional manifold.

I have been given the definition of the manifold:

A smooth manifold of dimension $m$ in $\mathbb{R}^n$ is a set $M$ with the following property: For each point $a \in M$ there is a function $F:\mathbb{R}^n \rightarrow \mathbb{R}^{n-m}$ which is regular on an open set $G$ containing $a$ and is such that

$M \cap G = (x:F(x) = 0) \cap G$

Now, I'm pretty sure this definition is all I need to answer this problem. I know that there must exist a function $F:\mathbb{R}^3 \rightarrow \mathbb{R}^1$ which is regular (although I'm not exactly sure how to show a function is regular: do I simply show it is one-to-one?). I just can't really wrap my head around what this F function is or how to create one to show somethings is an m-dimensional manifold. My intuition here really stinks.

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I think, by regular on G they mean dF is surjective on G.

Then they want: 1)to find F s.t. $F^{-1}(0)\cap G=M\cap G$ 2)0 is a regular value of F.

Note that the above "definition" is really a theorem: The smooth preimage of a regular value, is a smooth manifold.