Show there can be no co-ordinate patch at this point I am attempting to prove that the subset of $\mathbf{R}^3$ satisfying $x^2 + z^2 = y^2$ is not a surface, where

a surface is a subset of $\mathbf{R}^3$ for every point in which there is a co-ordinate patch whose image contains that point and is contained in the subset

and 

a co-ordinate patch is a smooth, injective, regular function from an open region of $\mathbf{R}^2$ to $\mathbf{R}^3$.

Having seen this visualized, I know this space is composed of two infinite cones joined at their tips (that is, at the origin). Intuitively, I then know that this is the problematic point that stops the entire set being a surface. However, I am having trouble constructing a reason why there cannot be a co-ordinate patch containing the origin; that is, showing which of the conditions on a co-ordinate patch cannot be satisfied, and why.
My attempt: If $\bar{x} : D \to \mathbf{R}^3$ were such a co-ordinate patch, I feel that because $D$ is open and $\bar{x}$ is continuous that there would necessarily be a "jump" from one of the cones composing the space to the other. As in, there would exist some point on the cone with positive $y$ and some point on the other cone with negative $y$ whose inputs in $D$ are close but which are obviously not close on the space.
Any help is greatly appreciated. Thank you in advance.
 A: For this one, there's a particularly easy argument. Suppose that the double-cone is called $D$, and that $U \subset D$ is the domain of a coordinate patch 
$$
p : U \to \Bbb R^2,
$$
hence there's an open set $W$ in 3-space with $W \cap D = U$. 
Within $W$, there's an open ball $W'$ centered at the origin. Let $U' = U \cap W'$. Then $U'$ is also the domain of a (smaller) coordinate patch. 
By changing coordinates in $\Bbb R^2$, we can assume that $p(0,0,0) = (0,0)$. 
Take a small open ball $B$ around the origin in $\Bbb R^2$. The preimage $Q = p^{-1}(B)$ is an open set in $D$, and contains the origin of 3-space. So $p|Q$, the restriction of $p$ to the set $Q$, is a homeomorphism from an open set in $D$ containing the origin to an open ball in $R^2$ containing the origin. 
That is to say $Q$ and $p(Q)$ are homeomorphic, and $p$ is a homeomorphism between them. (Important fact 1)
$Q$ is an open set of $D$, hence the intersection of an open set $U$ in 3-space with $D$. $U$ contains an open ball around the origin (because $(0,0,0) \in U$ and $U$ is open), hence it contains points in both the upper and lower cones of $D$, i.e., points with $y > 0$ and others with $y < 0$. (Important fact 2). 
Now let's look at 
$$
H = Q - \{(0,0,0)\}
$$
On the one hand, it's homeomorphic to $p(Q) - \{(0,0)\}$, which is a punctured disc, and is therefore connected (indeed, path connected!) 
On the other hand, $H$ consists of some portion of the double cone, but NOT the origin, the only point of $D$ with $y = 0$. So any path from a point in the upper-cone $(y > 0)$ to a point in the lower-cone $(y < 0)$ must be discontinuous (via the intermediate value theorem, if you like). Such pairs of points exist by important fact 2. 
Short form: If there were a patch, it'd make a disk homeomorphic to a little part of $D$ containing the origin AND parts of both cones. Removing the origin from both sides leads to something disconnected in the preimage, but connected in the image. THat's a contradiction. 
A: Hint
Call $C=\{x^2+z^2=y^2\}$ the cone. I introduce also $C_\pm=C\cap\{\pm y\geq0\}$.
Assume that there is $\phi\in C^1(D,C)$ with $D\subset \mathbb R^2$ open and connected, $0\in D$ and $\phi(0)=0$.
We prove that $\phi(D)$ is entirely contained in either $C_+$ or $C_-$. Assume the contrary. Then $\phi(D)\setminus\{0\}$ is not connected. But
$$
\phi(D)\setminus\{0\} = \phi(D\setminus\{0\})
$$
has to be connected because it is the image of the connected set $D\setminus\{0\}$.
So now you have $\phi\in C^1(D,C_+)$. Can you see what goes wrong with the derivatives of $\phi$ at the origin?
