Define $F(x,y,u,v)= 3x^2-y^2+u^2+4uv+v^2$ $G(x,y,u,v)=x^2-y^2+2uv$
Show that there is no open set in the $(u,v)$ plane such that $(F,G)=(0,0)$ defines $x$ and $y$ in terms of $u$ and $v$.
If (F,G) is equal to say (9,-3) you can just apply the Implicit function theorem and show that in a neighborhood of (1,1) $x$ and $y$ are defined in terms of $u$ and $v$. But this question seems to imply that some part of the assumptions must be necessary for such functions to exist?
I believe that since the partials exist and are continuous the determinant of $\pmatrix{ \frac{\partial F}{\partial x}&\frac{\partial F}{\partial y}\cr \frac{\partial G}{\partial x}&\frac{\partial G}{\partial y} }$ must be non-zero in order for x and y to be implicitly defined on an open set near any point (u,v) but since the above conditions require x=y=0 the determinant of the above matrix is =0.
I have not found this in an analysis text but this paper http://www.u.arizona.edu/~nlazzati/Courses/Math519/Notes/Note%203.pdf claims it is necessary.