If $\displaystyle f(x, y, z) = 0$, then [MATH%
first think in two dimensions. $\displaystyle f(x,y)=0$ means $\displaystyle y= \phi(x)$ although it may not always be possible to express $\displaystyle y$ as an explicit function of x.
similarly $\displaystyle f(x,y,z)=0$ means $\displaystyle z=\phi(x,y)$ $\displaystyle (\text{ or } y= g(z,x))$. Again it may not always be possible to express $\displaystyle z$ explicitly as a function of $\displaystyle x$ and $\displaystyle y$.
Geometrically $\displaystyle f(x,y,x)$ is a function of 3 independant variable. there is no constraint on the values of $\displaystyle x,y,z$(except that these must lie in the domain of the definition of the function). if you set $\displaystyle f(x,y,z)$ equal to zero you have effectively put 1 constraint, that is, if you choose arbitrary values of $\displaystyle x \text{ and } y$ then the value of $\displaystyle z$ should be so that the function $\displaystyle f$ must become zero. now since $\displaystyle z$ becomes fixed when $\displaystyle (x,y)$ is a point chosen(anywhere in the domain) on $\displaystyle x$-$\displaystyle y$ plane, the equation$\displaystyle f(x,y,z)=0$ is that of a surface.
For the sake of understanding suppose we could write $\displaystyle z=g(x,y)$ we have $\displaystyle dz= \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial y}dy$.
if $\displaystyle z$ is a constant then $\displaystyle 0=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y} \frac{\partial y}{\partial x}$
which on using $\displaystyle \frac{\partial g}{\partial x}=\frac{1}{\frac{\partial x}{\partial g}}$ we have,
$\displaystyle \frac{\partial y}{\partial x} \frac{\partial g}{\partial y} \frac{\partial x}{\partial g}=-1$
The thought that $\displaystyle f(x,y,z(x,y)) = 0$ came into my mind as well, but I couldn't go this far. What I was doing was that I try to differentiate with respect to $\displaystyle x$ and get $\displaystyle \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 0$ ...
I never have thought that $\displaystyle dz = 0$
I really appreciate your help.