Determine linear map so that equation is correct Let $f:\mathbb R^3→\mathbb R^2$, $(x,y,z)^T\mapsto(x^2y-5z,2x+4yz-3z^3)^T$,
and $(x_0, y_0, z_0)^T=(-1,0,1)^T$.
I need to determine the linear map $A∈L(\mathbb R^3,\mathbb R^2)\cong\mathbb R^{2×3}$ so that$$
f \begin{pmatrix} x \\ y \\ z \end{pmatrix}=f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}+A\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}+\left\|\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}\right\| ε\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}
$$
with $\lim\limits_{\vec h \to 0} ε(\vec h)=0$.
How can this be done? I don't know how to start (except for calculating $f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}$).
$f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}=\begin{pmatrix} -5 \\ -5 \end{pmatrix}$
$A=\begin{pmatrix} 0 && 1 && -5\\ 2 && 4 && -9 \end{pmatrix}$
$\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}=\begin{pmatrix} x+1 \\ y \\ z-1 \end{pmatrix}$
Therefore:
$f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}+A\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}=\begin{pmatrix}y-5z \\ 6+2x+4y-9z\end{pmatrix}$

Guide:
Use Taylor series expansion.
Let $(x,y,z) = (x_1, x_2, x_3)$ and $f(x,y,z)=(w_1,w_2)$.
Compute $A_{ij} = \frac{\partial w_i}{\partial x_j}$ and evaluate at $(x_0, y_0, z_0)$.
Edit:
For checking, I just drop higher order terms.
\begin{align}
x^2y-5z &= [(x-1)+1]^2y-5z \\
&\approx [2(x-1)+1]y - 5z \\
&\approx y-5z\\
2x+4yz-3z^3&= 2x+4y[(z-1)+1]-3[(z-1)+1]^3\\
&\approx2x+4y-3[3(z-1)+1]\\
&\approx 2x+4y-9z+6
\end{align}
