Partial derivative chain rule?

No, the chain rule does not involve such a minus sign.
Why are you asking?

 

It is true that IF the 3 variables x,y,z are related together in a sufficiently nice condition G(x,y,z)=0, THEN, we may solve for, in a region about a solution point, one of the variables in terms of the other two, i.e, functions X(y,z), Y(x,z), Z(x,y) exist, so that within that region we have that G(X(y,z),y,z)=G(x,Y(x,z),z)=G(x,y,Z(x,y)=0 identically.

In these cases, it is true that we have the counter-intuitive result:
[tex]\frac{\partial{Y}}{\partial{x}}\frac{\partial{X}}{\partial{z}}\frac{\partial{Z}}{\partial{y}}=-1[/tex]

 

To give a simple example.
Consider the function G(x,y,z)=x+y+z

Now, the condition G(x,y,z)=0 gives rise to the equation x+y+z=0
We may now form three separate function definitions:
X(y,z)=-y-z
Y(x,z)=-x-z
Z(x,y)=-x-y

We have now that:
G(X(y,z),y,z)=-y-z+y+z=0, i.e, the condition G=0 is satisfied IDENTICALLY, for all choices of y and z.
Similarly with the other two substitutions.

We see that in this case, that we have:
[tex]\frac{\partial{Y}}{\partial{x}}=\frac{\partial{X}}{\partial{z}}=\frac{\partial{Z}}{\partial{y}}=-1[/tex]
and therefore,
[tex]\frac{\partial{Y}}{\partial{x}}\frac{\partial{X}}{\partial{z}}\frac{\partial{Z}}{\partial{y}}-1[/tex]

 

In that specific case, the equation is true but it is NOT "the chain rule". Your initial post implied that you were offering this as a general formula derived from the chain rule.

 

"w ( x, y, z ). We can then derive z ( x, y, w )."
This is meaningless.