Confusion in applying the Implicit function theorem

Question

Consider the following equations $$\begin{cases} 2(x^2+y^2)-z^2=0\\ x+y+z-2=0\end{cases}$$Prove that the above system of equations defines a unique function $\phi: z\mapsto (x(z),y(z))$, from a neighborhood of $z=2$ to a neighbor hood $V$ of $(1,-1)$ and $\phi\in C^1$ on $U$.

My idea is to use The implicit function theorem
Now I have to check the condition to apply this theorem!
First, let set $F(x,y,z)=2(x^2+y^2)-z^2$ and $G(x,y,z)=x+y+z-2$. Obviously, $F,G\in C^1$ on $R^3$, $F(1,-1,2)=G(1,-1,2)=0$.
Also, $D_zF(1,-1,2)=-4\neq 0, D_zG(1,-1,2)=1\neq 0$.
According to the Implicit function theorem, there exits a unique $z=f(x,y)$ defined for $(x,y)$ near $(1,-1)$ s.t $F(x,y,z)=0$ and a unique $z=g(x,y)$ defined for $(x,y)$ near $(1,-1)$ s.t $G(x,y,z)=0$
Does this imply there is a unique function $\phi: z\mapsto (x(z),y(z))$, from a neighborhood of $z=2$ to a neighbor hood $V$ of $(1,-1)$ and $\phi\in C^1$ on $U$?

Answer 1

No, you must check that the $2\times 2$ Jacobian matrix $\dfrac{\partial (F,G)}{\partial (x,y)}$ is invertible at the given point $(1,-1,2)$. With regard to your final question, read the statement of the Implicit Function Theorem very carefully. What specific question do you have?

Answer 2

In this particular case you can solve the system explicitly. Namely $$ z^2+4xy=2(x+y)^2=2(2-z)^2$$ Thus $$xy={1\over 4}z^2-2z+2,\ x+y=2-z$$ By the Vieta formulas $x,y$ are solutions of the quadratic equation $$ u^2-(2-z)u+\left ({1\over 4}z^2-2z+2\right )=0$$ The discriminant is equal $4z -4.$ Therefore $$x,y ={1\over 2}[2-z\pm 2\sqrt{z-1}]$$ Taking into account the assumptions $x(2)=1$ and $y(2)=-1$ gives $$x=1-{z\over 2}+\sqrt{z-1},\ y=1-{z\over 2}-\sqrt{z-1}$$