I didn't really understand how implicit multi-variable functions are derived; I thought of another method which may fit and may not; suppose we have $xy^2z^3=8$ and we want to derive it; it is the same as $xy^2z^3 -8 =0$
or $ F(x,y,z) = 0 $
hence $ F(x,y,z) = xy^2z^3 -8 $
Therefore $\frac{\partial F}{\partial x} $, $\frac{\partial F}{\partial y} $, $\frac{\partial F}{\partial z} $ are in fact the partial derivatives of the implicit function that we sought. The three are also the coordinates of the normal at a certain point and of the gradient.
makes sense?
No. You can't do it like this.
There is a 3D surface given by the equation $F(x,y,z)=0$. You want to know, if you can describe this surface as a function $z=z(x,y)$ (for example). The implicit function theorem gives you an answer, when you can do it. So, the implicit function is the function $z$ not the function $F$ ($F$ just gives you the surface). It is implicit, because it is given by an equation ($F=0$) not by explicit formula ($z(x,y)=...$).
Then you have $$F(x,y,z(x,y))=0,$$ what you can partially differentiate (when?) with respect to $x$ and get $$\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0,$$ which gives you the formula for $\frac{\partial z}{\partial x}$. This is just the basic concept of course.