How to show this equation system does not define an implicit function? I am trying to show that the following equation system:
$$e^u+e^v+2-4e^{x^2-y}=0$$
$$\sin u+xu-v^4-y=0$$
Does not define $(y,x)$ as a function of $(u,v)$ implicitly around $(0,0,0,0)$. 
Attempt What I tried to do was find $u,v$ such that in a small radius of (0,0,0,0) the equation would have no solution. My best guess was $u=0, v=R$ (where $R =$ radius), but I had difficulty showing that the system then has no solution.
Help would be appreciated, and thanks in advance!
 A: You have the right idea. Setting $u=0$ simplifies the system to $e^v+3=4e^{x^2-y}$, $y=-v^4$. The key observation is that the exponent $x^2-y$ cannot be negative, and this is a problem for the first equation when $v<0$. 
A: The implicit function theorem guarantees you a nice solution $x=x(u,v)$, $\ y=y(u,v)$ with $x(0,0)=0$, $\ y(0,0)=0$ if a certain "technical condition" is satisfied, but has nothing to say if this condition is not satisfied (which is the case here).
On the other hand we can go and solve the system for $x$ and $y$ explicitly. From the second equation we deduce $y=x u+\sin u-v^4$, and plugging this into the first equation gives
$$e^u+e^v+2=4\exp(x^2-x u-\sin u+v^4)$$
or
$$x^2 - xu -\sin u +v^4-\log\Bigl(1+{e^u-1\over4}+{e^v-1\over4}\Bigr)=0\ .$$
This is a quadratic equation for $x$ which you can solve explicitly, but it might have  complex solutions for some $(u,v)$ near $(0,0)$. Maybe you don't allow this. 
To see what happens we retain only the  terms of order $\leq2$ in $u$ and $v$. We then have the equation
$$x^2-x u-{5\over4}u-{1\over4}v-{1\over32}(3u^2-2u v+3v^2)=0$$
with the formal solutions
$$x={1\over2}\Bigl(u\pm\sqrt{5u+v+{\rm quadratic\ terms}}\Bigr)\ .$$
Here the radicand is negative on one side of a quadric going through the origin of the $(u,v)$-plane. This implies that in the immediate neighborhood of $(0,0)$ there are points $(u,v)$ to which correspond no real $(x,y)$ solving the given system of equations.
