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?

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,
$$\nabla f \cdot \langle x-x_0,y-y_0,z-z_0 \rangle=0$$
