Tangent plane of a surface

First find out the two directional vectors:

$$\begin{align*} T_x &= \begin{bmatrix} 1 \\ 0 \\ f_x(x_0, y_0) \end{bmatrix}\\ T_y &= \begin{bmatrix} 0 \\ 1 \\ f_y(x_0, y_0) \end{bmatrix} \end{align*}$$

Then we have the normal vector of the tangent plane:

$$N = T_x \times T_y = \begin{bmatrix} -f_x(x_0, y_0)\\ -f_y(x_0, y_0)\\ 1 \end{bmatrix}$$

With $N$ in our pocket, we can easily get the equation for the tangent plane:

$$\begin{bmatrix} -f_x(x_0, y_0)\\ -f_y(x_0, y_0)\\ 1 \end{bmatrix} \cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} -f_x(x_0, y_0)\\ -f_y(x_0, y_0)\\ 1 \end{bmatrix} \cdot \begin{bmatrix} x_0\\ y_0\\ z_0 \end{bmatrix}$$

Or, equivalently:

$$\Delta z = f_x(x_0, y_0)\Delta x + f_y(x_0, y_0)\Delta y$$