Confusion about implicit differentiation and chain rule This silly question comes in two parts. First, I wanted to solve problem 58 in chapter 14.5 of Stewart, 8th edition, which states: Suppose that the equation $F(x,y,z) = 0$ implicitly defines each of the three variables x, y, and z as functions of the other two. In other words $z = f(x,y), y = g(x,z), x = h(y,z)$. If $F$ is differentiable and $F_x$, $F_y$, and $F_z$ are all non zero, show that
$$\frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = -1$$
Here's my solution which might be incorrect: I just write out $F_i$ for $i = x,y,z$ and you get for example, 
$$F_x = F_x \frac{\partial x}{\partial x}+F_y\frac{\partial y}{\partial x}+ F_z \frac{\partial z}{\partial x}$$
and then the first term is the same as what's on the left so you can solve for $\frac{\partial y}{\partial z}$. You do this for the other two variables and then multiply them together to get 
$$\frac{-F_x F_y F_z}{F_x F_y F_z}$$ 
(after some rearrangement) and then since the order of the derivatives doesn't matter, we're done. 
Ok, my question is, this doesn't seem to be entirely consistent with what was done on page 943, the last boxed equation of the section (8th edition of Stewart). Because here they have a function $F(x,y,f(x,y)) = 0$ which defines the variable $z=f(x,y)$. They want to find $\frac{\partial z}{ \partial x}$, so they write out $F_x$ and set it equal to zero like this: 
$$F_x \frac{\partial x}{ \partial x} + F_y \frac{\partial y }{\partial x} +F_z \frac{\partial z}{ \partial x} = 0$$
And then say that $\frac {\partial z}{\partial x} = -\frac{F_x}{F_z}$
Maybe I'm misunderstanding how to write out $F_x$?
Is my solution to the problem correct? If so, how are these things compatible? 
POSSIBLE ANSWER: I think I know the answer. What they are setting equal to zero is not the partial derivative of F(x,f(x,y)), but the total derivative with respect to $x$. 
My solution to problem 58 seems slightly wrong then...  
 A: Assume that at $p=(x_0,y_0,z_0)$ we have $$F(p)=0,\quad a:=F_x(p)\ne0,\quad b:=F_y(p)\ne0,\quad c:=F_z(p)\ne0\ .$$
Then there is a 3D window $W$ with center $p$ such that within $W$ the surface $F(x,y,z)=0$ can be viewed as  graph over a coordinate plane in three possible ways: There are $C^1$-functions $f$, $g$, $h$ in the indicated variables, such that within $W$ the equation $F(x,y,z)=0$ is equivalent with each of the following:
$$z=f(x,y),\quad x=g(y,z),\quad y=h(z,x)\ .$$
Therefore one has, e.g.,
$$F\bigl(x,y,f(x,y)\bigr)=0\qquad\forall\>(x,y)\in W'\ ,$$
and this implies, by the chain rule,
$$F_x\bigl(x,y,f(x,y)\bigr)+F_z\bigl(x,y,f(x,y)\bigr)f_x(x,y)\equiv0\ .$$
Evaluating this at $(x_0,y_0)$, and noting that $f(x_0,y_0)=z_0$, we obtain $a+c f_x(x_0,y_0)=0$, or
$$f_x(x_0,y_0)=-{a\over c}\ .$$
In the same way (note the cyclic symmetry) one derives
$$g_y(y_0,z_0)=-{b\over a},\qquad h_z(z_0,x_0)=-{c\over b}\ .$$
Multiplying the last three equations together we therefore obtain
$$f_x(x_0,y_0)g_y(y_0,z_0)h_z(z_0,x_0)=-1\ ,$$
and this can, in a somewhat sloppy way, be condensed to
$${\partial f\over\partial x}{\partial g\over\partial y}{\partial h\over\partial z}=-1\ .$$
Note that there is a second such formula, "going the other way around".
