I'm attempting to show that, given three variable $x,y$ and $z$, if they are related by the condition $F(x,y,z)=constant$ then the following reciprocity relation holds $$\frac{\partial x}{\partial y}=\frac{1}{\frac{\partial y}{\partial x}}$$ and similarly for $x$ and $z$, and $y$ and $z$.
From the condition $F(x,y,z)=constant$ it is clear that the following relations between the variables $x,y$ and $z$ can be constructed $$x=x(y,z)\;,\qquad y=y(x,z)\;,\qquad z=z(x,y)$$ and as such $$dx=\frac{\partial x}{\partial y}dy+\frac{\partial x}{\partial z}dz \\ dy=\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz \\ dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$ To derive the reciprocal relations from these differentials do we note that (for the first one that I put above) two of the variables must be independent and so we choose them to be $y$ and $z$. It follows that $$dy=\frac{\partial y}{\partial x}\left(\frac{\partial x}{\partial y}dy+\frac{\partial x}{\partial z}dz\right)+\frac{\partial y}{\partial z}dz\\ \Rightarrow \left(1-\frac{\partial y}{\partial x}\frac{\partial x}{\partial y}\right)dy=\left(\frac{\partial y}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial y}{\partial z}\right)dz$$ For this to be true $\forall y,z$ we must have that the coefficients on both sides of the equation are zero (as $y$ and $z$ are assumed independent) and this implies that $$1-\frac{\partial y}{\partial x}\frac{\partial x}{\partial y}=0\;\;\Rightarrow\;\;\frac{\partial x}{\partial y}=\frac{1}{\frac{\partial y}{\partial x}}$$
Would this be a correct analysis?