The implicit function theorem The problem I have is that the point $( 1,1,1,1)$ satisfies the equations $x+y^2-u-v=0$ , $xu-yv^2=0$ . Show that this defines $x , y$ as functions of $u,v$ in a neighbourhood of $(u,v)=(1,1)$ . The idea comes from the implicit function theorem but can somebody help me with that ? 
 A: Define $F: \Bbb R^4 \to \Bbb R^2$ by $$F(x,y,u,v) = (x+y²-u-v,xu-yv^2).$$We have $F(x,y,1,1) = (x+y^2-2.x-y)$. If we want $F(x,y,1,1)=(0,0)$, then we must have $x = y$ from the second component, and $x²+x-2 = 0$ from the first one. Solving gives $$x = \frac{-1\pm \sqrt{1+4\cdot 2}}{2} = \frac{-1\pm 3}{2} = 1 \text{ or }-2.$$Since $F(1,1,1,1) = (0,0)$ and $F(-2,-2,1,1) = (0,0)$, this does not define $x$ and $y$ as functions of $u$ and $v$.
You could compute $$DF(x,y,u,v) = \begin{bmatrix} 1 & 2y & -1 
& -1 \\ u & -v² & x & -2yv\end{bmatrix} \implies DF(x,y,1,1) = \begin{bmatrix} 1 & 2y & -1 
& -1 \\ 1 & -1 & x & -2y\end{bmatrix}, $$which does not allow us to conclude anything. If we had found unique $x_0$ and $y_0$ satisfying $F(x_0,y_0,1,1)$, then we could find a map $(x,y)\colon U \to V \subseteq \Bbb R^2$, for $U$ an open neighbourhood of $(1,1)$ in $\Bbb R²$ and $V$ an open neighbourhood of $(x_0,y_0)$, such that $F(x(u,v),y(u,v),u,v) = 0$ for all $(u,v) \in U$, if the block $$\begin{bmatrix} -1 & -1 \\ x_0 & -2y_0\end{bmatrix}$$were non-singular. But this is not the case.

For the edited problem: around $(1,1,1,1)$ we can take $x$ and $y$ in terms of $u$ and $v$, since by our previous computations we have $$ DF(1,1,1,1) = \begin{bmatrix} 1 & 2 & -1 
& -1 \\ 1 & -1 & 1 & -2\end{bmatrix}$$and the block $$\begin{bmatrix} -1 & -1 \\ 1  & -2\end{bmatrix}$$is non-singular.
