Tangent plane of the surface: $z-g(x,y)=0$ in $(x_0, y_0, z_0)$ How can I determine the equation of tangent plane of the surface:
$$z-g(x,y)=0$$
in the point:
$$(x_0, y_0, z_0)$$
in terms of implicit derivatives?
 A: Let be $f:\mathbb R^n\times\mathbb R^m\to\mathbb R^m$ and $c\in\mathbb R^m$ a regular value of $f$. Consider $(a,b)\in f^{-1}(c)$. 
The implicit function theorem gives an implicit function $\phi: U\to V$ where $U$ is a neighbourhood of $a$ and $V$ is a neighbourhood of $b$. 
Now you can define a parametrization of $f^{-1}(c)\cap (U\times V)$ by
$$
\Phi:U\to f^{-1}(c)\cap(U\times V),~\Phi(\alpha)=\begin{pmatrix}\alpha\\\phi(\alpha)\end{pmatrix}
$$
Consider $$
J_\Phi(\alpha_0)=\begin{pmatrix}I_m\\J_\phi(\alpha_0)\end{pmatrix},
$$ 
where $I_m$ is the identity matrix on $\mathbb R^m$, $J_\Phi(\alpha_0)$ is 
the Jacobian matrix of $\Psi$ at $\alpha_0$ and $J_\phi(\alpha_0)$ is the Jacobian matrix of $\phi$ at $\alpha_0$.
Since $\Phi$ is a parametrization, the columns of $J_\Phi(\alpha_0)$ forms a base of the Tangent space of $\Psi(U)=f^{-1}(c)\cap(U\times V)$ at $\Psi(\alpha_0)$.
In your case $g$ is the implicit function of $f(x,y,z):=z-g(x,y)$, $c=0$ is a regular value of $f$ and
$$
\begin{pmatrix}I_2\\J_g(x_0,y_0)\end{pmatrix}=\begin{pmatrix}1&0\\0&1\\\partial_xg(x_0,y_0)& \partial_yg(x_0,y_0)\end{pmatrix}.
$$
Therefore your tangent space of $z-g(x,y)=f(x,y,z)=0$ at $(x_0,y_0,z_0)$ is 
the span of 
$$
\begin{pmatrix}1\\0\\\partial_xg(x_0,y_0)\end{pmatrix}
\text{ and }
\begin{pmatrix}0\\1\\ \partial_yg(x_0,y_0)\end{pmatrix}.
$$


But you don't need to use the implicit function theorem, because you can directly see that $$
\Psi(x,y)=\begin{pmatrix}x\\y\\g(x,y)\end{pmatrix}
$$
is a parametrization of your plane $z-g(x,y)=0$.
A: The gradient of a function is normal to its level curves. Proof:

Suppose $\vec r(t)=\langle x(t),y(t),z(t) \rangle$ parametrizes the curve $f(x,y,z)=c$, where $c \in \mathbb{R}$ is some constant. Then, 
$$f(x(t),y(t),z(t))=c$$
Since $\vec r(t)$ is on the curve. We also have,
$$\frac{d}{dt}f(x(t),y(t),z(t))=0$$
By the chain rule this is the same as,
$$x’(t) \cdot f_x+y’(t) \cdot f_y+z’(t) \cdot f_z=0$$
$$\vec r’(t) \cdot \nabla f=0$$
A normal vector by definition is orthogonal to a tangent vector. Hence $\nabla f=\langle f_x,f_y,f_z \rangle$ at some point $p$ describes a normal vector to $f(x,y,c)=c$ at that point $p$.

Let $f(x,y,z)=z-g(x,y)$ then because $\nabla f$ is a normal vector to the level curves of $f$ it will be a normal to $z-g(x,y)=0$. Hence the equation of the tangent plane of $z-g(x,y)=0$ at $(x_0,y_0,z_0)$ is,
$$\nabla f \cdot \langle x-x_0,y-y_0,z-z_0 \rangle=0$$
