Why is this set not a manifold? Set $M = \{ \, (x, y) : x^2 = y^2 \, \}$.  If for every point $(a, c)$ in $M$, there exists a neighborhood $U$ containing $(a, c)$ and function $\phi(x, y)$ such that:


*

*$\phi(x, y) = 0$ on $M \cap U$;

*The Jacobian matrix associated with $\phi$ has rank $1$ on $U$.  (In general, it does not have to be rank $1$.  But here the only choice is $1$.)


Then, $M$ is a manifold.  If the Jacobian matrix has ranks greater than $0$, then we have use $\phi$ to carry out the implicit function theorem, and construct a function such that $(x, y) = (x, f(x))$ on $M \cap U$.  But I don't know how to go in reverse; what is the contradiction if $M$ is a manifold?
An educated guess says that $(0, 0)$ is our trouble spot.  The function $\varphi(x, y) = x^2 - y^2$ equals $0$ on $M$.  But the Jacobian matrix has zero rank at $(0, 0)$.  So, we cannot use $\varphi$ to carry out the implicit function theorem...
 A: $x^2-y^2=0$ is equivalent to $(x-y)(x+y)=0$, it is equivalent to $x=y$ or $x=-y$. Thus it is the union of two lines of $R^2$ which intersects at $(0,0)$, it is not a manifold since you don't have a tangent space at $(0,0)$. 
You can also say that if it was a manifold, it would have been a $1$-dimensional manifold, but a connected neighborhood of $(0,0)$ can't be diffeomorphic to an interval, since if you remove $(0,0)$ from it, you have at least four connected components.
A: A 1-dimensional connected manifold has the property that when you remove any one of its points you get at most two connected components.
Prove this and then use it to show that your set is not a manifold.
A: $\phi$ has not just the purpose to indicate which points are in the manifold. It also gives a kind of “local compass”, telling you from each point in which unique direction the manifold continues, namely (one-dimensional case) in direction $\pm\nabla \phi$. Note that really only the direction is meaningful: the distance which you can go to stay in the manifold is in general just infinitesimal; so it's actually rather $\frac{\nabla\phi}{\|\nabla\phi\|}$ that's interesting†. But this is not defined if $\nabla\phi=0$ at some point – in this case it can't point you to any one direction where the manifold continues. In your  example, at $p=(0,0)$ there are actually two directions in which $M$ extends: $(1,1)$ and $(1,-1)$.
As you say, one way to grasp this problem is to say that it prevents you from applying the implicit function theorem.

†By the way, the space of these “local directions” forms itself a manifold structure.
