I am trying to solve a question and I am stuck on something.
Consider the equations:$$e^{x}+y^{2}z=2,\ \ xy+2y+\frac{1}{z^{2}}=3$$I should find (i) If there exists a neighborhood of (0,1,1) such that these equations define x,z as a differentiable function of y. (ii) same thing but x,y as a function of z.
So I defined $$G\left(x,y,z\right)=\begin{pmatrix}e^{x}+y^{2}z-2\\ xy+2y+\frac{1}{z^{2}}-3 \end{pmatrix}$$ and found out that $\begin{pmatrix}\frac{\partial G_{1}}{\partial x} & \frac{\partial G_{1}}{\partial z}\\ \frac{\partial G_{2}}{\partial x} & \frac{\partial G_{2}}{\partial z} \end{pmatrix}$ is invertible and thus, the function we want exists. For (ii) it came out that $\begin{pmatrix}\frac{\partial G_{1}}{\partial x} & \frac{\partial G_{1}}{\partial y}\\ \frac{\partial G_{2}}{\partial x} & \frac{\partial G_{2}}{\partial y} \end{pmatrix}$ is not invertible and so I can't use the implicit function theorem. How one goes on from here? I tried to show that if I fix $z=1$ I got couple of solutions to:$$e^{x}+y^{2}=2,\ \ xy+2y=2$$ In a every neighborhood of (0,1,1) but it didn't work. How can one prove a function of this sort exist or not?
There can be no such differentiable functions $x(z)$ and $y(z)$, since if they existed you could differentiate the identity $G \bigl( x(z),y(z),z \bigr) = (0,0)^T$ to get $$ \begin{pmatrix} e^x & 2yz \\ y & x+2 \end{pmatrix} \begin{pmatrix} x'(z) \\ y'(z) \end{pmatrix} + \begin{pmatrix} y^2 \\ -2/z^3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} , $$ which at the point $(0,1,1)$ becomes $$ \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x'(1) \\ y'(1) \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} , $$ a contradiction.