$\delta\to \partial$: Is this argument valid? Is the following reasoning valid?
Suppose $z=z(x,y)$, then
$$\delta z =\left({\partial z \over \partial x}\right)_y\,\,\,\delta x+\left({\partial z \over \partial y}\right)_x\,\,\,\delta y$$
Divide through by $\delta x$ but treating them as partial derivatives gives
$$\left({\partial z \over \partial x}\right)_z=\left({\partial z \over \partial x}\right)_y\,\,\,\left({\partial x \over \partial x}\right)_z+\left({\partial z \over \partial y}\right)_x\,\,\,\left({\partial y \over \partial x}\right)_z$$
giving
$$0=\left({\partial z \over \partial x}\right)_y+\left({\partial z \over \partial y}\right)_x\,\,\,\left({\partial y \over \partial x}\right)_z$$?
I am not too comfortable with changing the $\delta$ into $\partial$ step. Does it make sense?
Thank you. :)
 A: There's a lot of this out there: people in doubt if it's possible or not to divide these "infinitesimal" changes. Well, I'll give my best to explain this:
As Spivak says in his "Calculus on Manifolds", in a mathematically rigourous framework, this notion of given a function $f : \mathbb{R}^n \to\mathbb{R}$ expressing infinitesimal of $f$ as a combination of infinitesimal changes of the parameters is meaningless.
In fact, there is a mathematically rigorous version of $\operatorname{d}f$ which is called the exterior derivative of $f$. In truth, within this trully rigorous framework, we can write:
$$\operatorname{d}f_p=\sum_{i=1}^n D_if(p) \ \operatorname{d}x^i$$
Where $D_i f(p)$ is the $i$-th partial derivative of $f$ at $p$. Here we call $\operatorname{d}f_p$ exterior derivative of $f$ at the point $p \in \mathbb{R}^n$. The main point, is that using this kind of treatment, the $\operatorname{d}f_p$ and the $\operatorname{d}x^i$ are linear functionals, in other words, they're functions of vectors, elements of the dual space of the space of vectors. Division of vectors is undefined, and also division of linear functionals is also undefined, then you can't formally do this.
Why should you care about it ? Well, these notions are more precise than that notion of infinitesimals. In reallity, this captures the ideas pretty well: this $\operatorname{d}f_p$ is a function that given a displacement, gives you one linear approximation to the change in $f$ at that point. It's more precise, because you're never stating that the change is that, you're saying that the change is that up to some error of second order.
So, the short answer to your question is: no, to derive your result preciselly and more rigourously you can't simply divide by that "infinitesimal" change.
My sugestion is that you look on Apostol's Calculus Volume 2. It's very complete, very rigorous and yet has more examples than Spivak. Of course Spivak's even more complete and rigorous since it's a book of analysis instead of calculus, but to get started with the replacement of this intuitive approach to the fully rigorous approach, Apostol's a good choice. Also, he doesn't present the exterior derivative this way, but it'll get you started so that you can better understand how to do multivariable calculus without that kind of ticks. Also, the result you're trying to derive he derives there within a rigorous framework, so it's worthwile taking a look.
A: Yeah, you've got the right idea.  The final form is usually written something like
$ \left({\partial y}\over{\partial x}\right)_z
\left({\partial z}\over{\partial y}\right)_x
\left({\partial x}\over{\partial z}\right)_y = -1
$
You have a $\delta x$ and a $\delta y$ and you are picking them so that as they change, $z$ remains constant, i.e. $\delta z = 0$.  Once you have that settled, it all boils down to the usual way of taking the limit of $\delta y/\delta x$ to get the derivative $dy/dx$ or, in this case $\partial y/\partial x$.
