$f(x,y,z)=z^2x+e^z+y$, implicit function theorem

Question

$f: \mathbb{R}^3 \rightarrow \mathbb{R}$, $f(x,y,z)=z^2x+e^z+y$

1. Show that a neighboorhood $V$ of $(1,-1)$ in $\mathbb{R}^2$ and a continuous differentiable function $g:V \rightarrow \mathbb{R}$ with $g(1,-1)=0$ and $f(x,y,g(x,y))=0$ for $(x,y) \in V$ exists.

2. Calculate $D_1g(1,-1)$ and $D_2g(1,-1)$

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$\nabla f(x,y,z)=\begin{pmatrix} f_x\\ f_y\\ f_z \end{pmatrix}=\begin{pmatrix} z^2\\ 1\\ e^z+2zx \end{pmatrix}$

1. I just have to show that $f(1,-1,0)=0$ and $f_z(1,-1,0) \ne 0$, the implicit function theorem shows that it holds. $f(1,-1,0)=0^2+e^0-1=0$ and $f_z(1,-1,0)=e^0+0=1\neq0$
2. I know that I have to differentiate $0=xg(x,y)^2+e^{g(x,y)}+y$ with respect to $x$ and $y$. But how do I do that?

Answer 1

Hint:

By chain rule, \begin{align} 0 &=\frac {\partial}{\partial x}f(x, y, g(x, y)) \\&= \left( f_x, f_y, f_z\right) \cdot \left(1, 0, D_1g(x,y)\right)\end{align}

Thus $f_x(1,-1,0)+ f_z(1,-1,0)D_1g(1, -1)=0$.