Implicit differentiation of function For a function $$ f(xyz, x+y+z)=0 $$ where z is defined as an implicit function of x and y, prove that: $$ x(y-z)Zx' +y(z-x)Zy' = z(x-y)$$
Im stuck on this problem if anyone is willing to explain how to do it. I know we have to use the Chain Rule for it.
to clarify the notation of the question:
$$ Zx' = \frac{dz}{dx} =  -\frac{\frac{df}{dx}}{\frac{df}{dz}}$$
 A: The notation is crazy, here is how I interpret it: we have a function of two variables, $f(u,v)$. Each of $u,v$ is a function of three variables $x,y,z$; explicitly $u(x,y,z)=xyz$ and $v(x,y,z)=x+y+z$.  For some function $z(x,y)$ of two variables we have that $f(u(x,y,z(x,y)), v(x,y,z(x,y))=0$ for all $x,y$.  
I assume your notation is a definition, so that we can replace
$$Zx' \text{ by } -{\frac{\partial f}{\partial x}}/{\frac{\partial f}{\partial z}}$$
and 
$$Zy' \text{ by } -{\frac{\partial f}{\partial y}}/{\frac{\partial f}{\partial z}}.$$
[Note that your use of total derivatives is quite inappropriate, and even the partials are possibly confusing. My solution indicates what we mean by them here.]
If we do this then what we must prove is
$$
\textbf{ (*) }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }
x(y-z)\frac{\partial f}{\partial x}+
y(z-x)\frac{\partial f}{\partial y}+
z(x-y)\frac{\partial f}{\partial z}=0.  
$$
To see this we use the Chain Rule on 
$$f(u(x,y,z),v(x,y,z)),
\text{ where } u(x,y,z)=xyz \text{ and } v(x,y,z)=x+y+z.
$$
to get
$$
\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u} yz + \frac{\partial f}{\partial v},\text{ } 
\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u} zx + \frac{\partial f}{\partial v},\text{ } 
\frac{\partial f}{\partial z}=\frac{\partial f}{\partial u} xy + \frac{\partial f}{\partial v}.
$$
Now we just plug these into the LHS of $(*)$ and see that it reduces at once to $0$. 
