1. ## Implicit Function Problem

If $f(x, y, z) = 0$, then [MATH%

2. Originally Posted by armeros
If $f(x, y, z) = 0$, then $\frac{\partial z}{
\partial y} \frac{\partial y}{
\partial x} \frac{\partial x}{
\partial z} = -1$

Make sense out of this nonsense and prove.

I am totally clueless.

I think $x, y, z$ can be written as functions of others

$x(z(y)), y(x(z)), z(y(x))$.

Anyone has a clue ?

Thanks.
first think in two dimensions. $f(x,y)=0$ means $y= \phi(x)$ although it may not always be possible to express $y$ as an explicit function of x.
similarly $f(x,y,z)=0$ means $z=\phi(x,y)$ $(\text{ or } y= g(z,x))$. Again it may not always be possible to express $z$ explicitly as a function of $x$ and $y$.

Geometrically $f(x,y,x)$ is a function of 3 independant variable. there is no constraint on the values of $x,y,z$(except that these must lie in the domain of the definition of the function). if you set $f(x,y,z)$ equal to zero you have effectively put 1 constraint, that is, if you choose arbitrary values of $x \text{ and } y$ then the value of $z$ should be so that the function $f$ must become zero. now since $z$ becomes fixed when $(x,y)$ is a point chosen(anywhere in the domain) on $x$- $y$ plane, the equation $f(x,y,z)=0$ is that of a surface.

For the sake of understanding suppose we could write $z=g(x,y)$ we have $dz= \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial y}dy$.
if $z$ is a constant then $0=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y} \frac{\partial y}{\partial x}$
which on using $\frac{\partial g}{\partial x}=\frac{1}{\frac{\partial x}{\partial g}}$ we have,
$\frac{\partial y}{\partial x} \frac{\partial g}{\partial y} \frac{\partial x}{\partial g}=-1$

3. The thought that $f(x,y,z(x,y)) = 0$ 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 $x$ and get $\frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 0$ ...

I never have thought that $dz = 0$