Do we consider input variables as independent when computing partial derivatives using Chain Rule? I am trying to follow the example here.
Considering the first statement:
Let $x = x(t)$ and $y = y(t)$ be differentiable at $t$ and
suppose that $z = f(x, y)$ is differentiable at the point
$(x(t), y(t))$. Then $z = f(x(t), y(t))$ is differentiable at $t$
and $$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} $$
The part I have trouble grasping is the meaning of $\frac{\partial z}{\partial x}$. It should mean "how does z change if you only vary $x$ while holding $y$ constant?" But this seems to be a conundrum as you cannot vary $x$ alone because $x$ and $y$ are also related $y = y(x^{-1}(x))$. 
 A: Sorry, I didn't read the example carefully. The short answer is: The example is very badly written. The partial derivatives are those of $f$, not of $z$, and should be written accordingly, $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. There should be a clear distinction between the function $z(t)$ of one variable and the function $f(x,y)$ of two variables. If you can, dump this tutorial and use another one.
Especially in physics, function names are sometimes overloaded to represent both a function of one variable and a function of two variables, as in
$$
\frac{\mathrm dV}{\mathrm dt}=\frac{\partial V}{\partial x}\frac{\mathrm dx}{\mathrm dt}+\frac{\partial V}{\partial y}\frac{\mathrm dy}{\mathrm dt}\;.
$$
This is already confusing, but you can keep it clear in your mind if you know which instances of $V$ stand for a function of one variable (the one on the left) and which stand for a function of two variables (the ones on the right). But in your example, a new symbol $z$ is introduced in the context of substituting $x(t)$ and $y(t)$ for $x$ and $y$, so it would be predestined to use it to keep the distinction clear; only to then be treated as if it were synonymous with $f$, which raises the question why it was introduced as a separate symbol in the first place.

My original answer before I noticed that it said $\frac{\partial z}{\partial x}$ instead of $\frac{\partial f}{\partial x}$; I'll leave it here in case it's nevertheless helpful:
$f(x,y)$ exists independently of $x(t)$, $y(t)$ and $z$. It has two independent arguments, $x$ and $y$. It's slightly unfortunate that the example overloads these and also uses them to denote functions. The example is easier to understand if you give these functions different names.
Imagine $f(x,y)$ to be the height of a landscape above sea level, $x(t)$ and $y(t)$ your $x$ and $y$ coordinates at time $t$ as you go hiking, and $z(t)$ your height above sea level at time $t$. The landscape $f$ is there whether you hike in it or not, and we can form its partial derivatives. We happen to be using them here to determine your climbing rate while you're hiking, but tomorrow we might use them to calculate the speed and direction of water rushing down the incline.
