Proof of Multivariable Implicit Differentiation Formula If the equation $F(x,y,z)=0$ defines $z$ implicitly as a differentiable function of x and y, then by taking a partial derivative with respect to one of the independent variables (in this case x), you get
$\large F_x(x,y,z)\frac{\partial x}{\partial x}+F_y(x,y,z)\frac{\partial y}{\partial x}+F_z(x,y,z)\frac{\partial z}{\partial x}=0.$
Because dx/dx = 1 and dy/dx = 0, you can solve for the desired partial derivative:
$\large \frac{\partial z}{\partial x}=-\frac{F_x(x,y,z)}{F_z(x,y,z)} $
The bolded dy/dx = 0 is what I don't get. I mean, it makes sense that an independent variable doesn't change in response to another, but it doesn't seem very formal and I feel like there's more to it than that. So basically, is there a more formal or detailed explanation or is that all there is to it?
 A: Going from $F(x,y,z)=0$ to $$F_x(x,y,z)\frac{\partial x}{\partial x}+F_y(x,y,z)\frac{\partial y}{\partial x}+F_y(x,y,z)\frac{\partial z}{\partial x}=0\tag{A}$$
is a formal manipulation, we are just applying $\frac{\partial}{\partial x}$ to both sides of $F(x,y,z)=0$. Then we are assuming that $z$, at least locally, can be written as a (smooth) function of $x$ and $y$. If we wish to consider $\frac{\partial z}{\partial x}$, we may recall what a partial/directional derivative stands for: the rate of change of a smooth function along a direction. In particular $\frac{\partial z}{\partial x}$ accounts for what happens to $z=z(x,y)$ when we move from $(x,y)$ to $(x+\varepsilon,y)$. What happens to $y$ during this travel? Absolutely nothing, it does not change. So by assuming $z=z(x,y)$ and recalling the geometric meaning of $\frac{\partial}{\partial x}$ we get that $(A)$ implies 
$$ F_x(x,y,z)+F_z(x,y,z)\frac{\partial z}{\partial x}=0 \tag{B} $$
as wanted. Remark: we would have got $(B)$ also by assuming that $F$ is locally linear, and not by chance. Indeed, the differentiable functions are the functions which are well-approximated by their tangent plane at any point.
