How can we show the cone $x^2 +y^2 = z^2$ is not a smooth manifold? In our differential geometry class, as a preliminary we have used as the definition of a manifold the following: $M \subset \mathbb{R}^n$ is a $k$-manifold if for each point $p\in M$ there exists a neighborhood $U$ of $p$ and $I = \{i_1, \dotsc, i_k\} \subset \{1,\dotsc,n\}$ such that $U \cap M$ is the graph of a $C^\infty$ function $f: V \to \mathbb{R}^{I^c}$, where $V \subset \mathbb{R}^I$.
One problem we've been assigned is to show that the cone 
$$C := \{(x,y,z) \in \mathbb{R}^3 \mid x^2 + y^2 = z^2 \}$$
is not a smooth manifold.
I have not yet learned any general ways to show something is not a smooth manifold. I feel like there may be a more clean way based on the more abstract definition of a smooth manifold, but I need something that works with this more limited definition.
$C$ is a level set of the function $F(x,y,z) = x^2 + y^2 - z^2$. We note that $\nabla F(\mathbf{0}) = \mathbf{0}$, so the implicit function theorem doesn't allow us to solve out for one variable in terms of the others at $\mathbf{0}$. But I've never learned a converse to the implicit function theorem...how can we show that it is not possible for such a parameterization of $C$ to exist?
Hints are appreciated; no need to feed me the answer. Thanks.
 A: Another hint: If you remove a point from a connected topological manifold of dimension $>1$, the result is still connected.  Now, look at your quadric and try to figure out which point disconnects it. 
A: Hint. Is the graph of $x\mapsto |x|$ a smooth manifold? It can be described as $\{(x,z)\ |\ x^2=z^2, z\geq 0\}$.
Update on the request of the OP:
The graph is not a smooth manifold, because:


*

*it is not (near $0$) a graph of the variable $y$, because any of the reasons:
1) there are two values of $x$ for one $y$,
2) the set $V$ in your definition must be required be open (it is a mistake), and it can't be open in this case, because $y$ is nonnegative.

*it is not a $C^\infty$ graph of $x$ (certainly it is a graph of $x\mapsto |x|$ [thus can't be a graph of any other function of $x$] and certainly this function isn't $C^\infty$).
A: Let us use your "preliminary" definition of a $k$-manifold in ${\mathbb R}^n$. You are given the set
$$S:=\{(x,y,z)\in{\mathbb R}^3\ |\ f(x,y,z)=0\},\qquad f(x,y,z):=x^2+y^2-z^2\ .$$
As you have remarked, at any point ${\bf p}\in S$ with ${\bf p}\ne{\bf 0}$ one has $\nabla f({\bf p})\ne{\bf 0}$. By the implicit function theorem in the neighborhood of such ${\bf p}$ the set $S$ satisfies your manifold condition with $k=2$.
Remains the point ${\bf p}={\bf 0}\in S$, and we stick to $k=2$. Selecting $I=\{1,2\}$ we obtain for each $(x,y)$ near ${\bf 0}$, but $\ne{\bf 0}$, two values $z$ such that $f(x,y,z)=0$, namely $z=\pm\sqrt{x^2+y^2}$. It follows that this $I$ does not qualify. Trying $I=\{1,3\}$, i.e., $x$ and $z$ as independent variables, we obtain
$$y=\pm\sqrt{z^2-x^2}\ ,$$
which is even undefined in half of any neighborhood of $(0,0)$. Same thing with $I=\{2,3\}$. 
It follows that at ${\bf 0}$ the manifold character of $S$ is definitively defect.
