$S = \{(x,y,z) : F(x,y,z) = 0\}$
$F \in C^{1}$
$F_{z} \neq 0$
In the solution using the implicit function theorem they somehow reach $dS = \sqrt{1 + f_{x}^{2} + f_{y}^{2}}$ . and this is my question.
They do not explain too much how, something about plane colliding with the graph, but I fail to see the true meaning of that and how we reach the above equation from it. I only know that the tangent plane in a point in this scenario should be
$z-z_{0} = f'_{x}(x_{0},y_0) (x-x_{0}) + f'_{y}(x_{0},y_0) (y-y_{0}) $
Don't know what to do from here to get to the $dS$ equation.
I do not think the rest of the question is necessary because I only need to know how to reach the dS equation above from the given data to solve the problem.
If I understood correctly, you have $F$ and you want to determine $\sqrt{1 + f_{x}^{2} + f_{y}^{2}}$.
By differentiating with respect to $x$ and to $y$ the implicit equation $F(x,y,z)=0$ we obtain, respectively, $$F_x+F_z \cdot f_x=0\;,\; F_y+F_z \cdot f_y=0\implies f_x=-\frac{F_x}{F_z}\;,\; f_y=-\frac{F_y}{F_z},$$ where $z=f(x,y)$. Hence $$\sqrt{1 + f_{x}^{2} + f_{y}^{2}}=\frac{\sqrt{F_x^2+F_y^2+ F_z^2}}{|F_z|}.$$