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

Question

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

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

Can someone help me with this?

Answer 1

Show that $f(1,-1,0)=0$ and $f_z(1,-1,0) \ne 0$.The implicit-function - theorem then shows that 1. holds.

For $(x,y) \in V$ we then have

$$(*) \quad 0=xg(x,y)^2+e^{g(x,y)}+y.$$

Differentiate $(*)$ with respect to $x$, plug in $(x,y)=(1,-1)$, then you can compute $D_1g(1,-1).$

Differentiate $(*)$ with respect to $y$, plug in $(x,y)=(1,-1)$, then you can compute $D_2g(1,-1).$