Assumption in deriving implicit function theorem?

I'm have a bit of trouble understanding how dependence of variables work with implicit functions. This troubles me during the chain rule and specifically in this case, the implicit function theorem.

The way I have seen the IFT derived

Suppose we have a function $$F(x,y,z)=c$$, and we are finding $$\frac{\partial F}{\partial x}$$.

At this point, if I'm not mistaken, we asumme that $$x$$ and $$y$$ are "completely independent" and that $$z$$ can be expressed as a function of $$x$$ and $$y$$ locally. So $$F(x,y,f(x,y))=c$$

And then we differentate both sides with respect to $$x$$, using the chain rule on the left hand side $$\frac{\partial F}{\partial x}\cdot1+\frac{\partial F}{\partial y}\cdot0+\frac{\partial F}{\partial z}\cdot\frac{\partial z}{\partial x} =0$$

My question is regarding this:

Question. Why is it a reasonable assumption to make that $$x$$ and $$y$$ are completely independent? (Especially when they are related by the function $$F(x,y,z)=c$$) If this is false (i.e. $$x,y$$ are dependent), then how can we conclude $$\frac{\partial y}{\partial x} = 0$$ which seems to be required to derive the formula for the implicit function theorem?

My question also extends beyond just the implicit function theorem:

In general if we have a function $$z=f(x(u),y(u))$$ and we're differentiating with respect to $$x$$. Usually what I would do is $$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x}$$ by treating $$y$$ as a constant.

But aren't $$y$$ and $$x$$ related? Why is $$\frac{\partial y}{\partial x} = 0$$ if a change in $$x$$ suggests a change in $$u$$ and thus a change in $$y$$ (Since $$y$$ also depends on $$u$$)?