Necessary Conditons For Implicit Function The Inverse Function Theorem provides sufficient conditions to determine when a function is defined implicitly by a relation. I would like to know some ways to determine when no such function is defined.
Below is a link to a specific example and conjecture. 
https://math.stackexchange.com/questions/46750/how-to-prove-the-implicit-function-theorem-fails
 A: In the same spirit as Michael's answer: Over the complex variables, there is a theorem due to W.F. Osgood (published in his "Lehrbuch der Funktionentheorie", 2. Aufl., Bd. 2, Parte 1, Leipzig 1929) about solvability of $w=f(z)$ for a system of holomorphic functions on a neighborhood of a point $a \in \mathbb{C}^n$, which is an isolated point of the set $\{z:f(z)=b:=f(a)\}$. This is nicely discussed in B.V. Shabat's book  "Complex analysis" (part II, section 14, item 44-although I am not sure if this is included in the English translation of the book)-in terms of resultants and Weierstrass' Preparation Theorem. Shabat also refers to: M. Herve, "Several Complex Variables. Local Theory", Oxford 1963.
And there is some information (on the topic of failure of the implicit function theorem)  in the book by Aizenberg and Yuzhakov on residue theorems.
A: A natural approach would be to classify such singular points by the deficiency of the rank of the Jacobian. If the deficiency is one, you can solve for all but one of the variables and reduce the problem to a scalar equation. The rest is then quite straightforward: The equation f(x)=y, with f(0)=0 is solvable for x in a neighborhood of 0 if the leading term in the Taylor expansion of f is odd; it is not always solvable if the leading term is even. If the deficiency in the rank of the Jacobian is two, you end up with a system of two equations, generally quadratic at leading order. Discussing the solvability of such a system is still a manageable task.
A: In this context, a remarkable phenomenon to consider is also lack of uniqueness, that may be considered an instance of bifurcation.
A: Just a comment about the link you posted: G=0 implies $4uv=2y^2-2x^2$ so when you substitute in $F=0$ you simply get $x^2+y^2+u^2+v^2=0$ which can be only satisfied by $x=y=u=v=0$ (if you work in $\mathbb{R}^4$...)
