Multivariable Chain Rule Formula doesn't make sense to me Consider $z=f(x(t),y(t))$, then its chain rule formula is:
$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$
What I cannot make sense of, are the terms $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$
$\frac{\partial f}{\partial x}$ is defined as the rate of change of $f$ with respect to $x$, when $y$ is held constant.
Now, in our original definition, both $x$ and $y$ depend on $t$. So if we hold $y$ constant, we'll have to hold $t$ constant, and thus $x$ will automatically be held constant. How, then, can we compute $\frac{\partial f}{\partial x}$ if we cannot hold only $y$ as constant?
 A: What is confusing you is that the letters $x$ and $y$ are used for two purposes: (a) As names of the functions $x(t), y(t)$; (b) As names of the "first argument" and "second argument" of $f$. Sadly, this notation is very usual in mathematics, and there is no suitable alternative.
If I could convince to write partial derivatives differently, e.g. instead of $\frac{\partial f}{\partial x}$ you write $f_1'$ ("derivative on the first parameter"), and instead of $\frac{\partial f}{\partial y}$ you write $f_2'$ ("derivative on the second parameter"), then the formula becomes a lot more palatable:
$$z'(t)=\frac{d}{dt}f(x(t),y(t))=f_1'(x(t),y(t))x'(t)+f_2'(x(t),y(t))y'(t)$$
(I also write $x'(t)=\frac{dx}{dt}, y'(t)=\frac{dy}{dt}, z'(t)=\frac{dz}{dt}$, if that helps.)
Now it is obvious we are in no way modifying $x(t)$ or $y(t)$ when calculating $f_1'$ and $f_2'$. Instead, we are keeping constant the second argument of $f$/the first argument of $f$, respectively. After having calculated $f_1'$ and $f_2'$ - both end up as functions of two arguments - you will substitute those arguments with $x(t), y(t)$.
Sadly, this notation I used above is completely non-standard, and for historical reasons we are stuck with the notation that is confusing for you.
A: Maybe an example would help. Say we take $z=x^2y$ where as you said $x,y$ are functions of $t$ so we will define them as $x=t^2$ and $y=t^3$.
Our formula is:
$$\frac{dz}{dt}=\frac{\partial z}{\partial x} \frac{dx}{dt}+\frac{\partial z}{\partial y} \frac{dy}{dt}$$
Which is then:
$$\frac{dz}{dt}=(2xy)(2t)+(x^2)(3t^2)$$
Subbing for $x,y$:
$$\frac{dz}{dt}=(2t^5)(2t)+(t^4)(3t^2)$$
$$\frac{dz}{dt}=7t^6$$
I hope this makes sense. I think the way you are getting confused is with the $f$ and $z$ and what you are differentiating with respect to. If you follow this example closely you should see!
A: I think the confusion comes from thinking about actually changing the values of the variables $x$ and $y$ when calculating $\frac{df}{dx}$ and $\frac{df}{dy}$. Just think of it as exploring what would happen to the function $f$ if $x$ or $y$ were to change (independently) by a small amount. Likewise for $\frac{dy}{dt}$ and $\frac{dx}{dt}$. Now, $x$ or $y$ cannot ever change independently since they both depend on t. But since calculating the derivatives, we now have information about how a small change in $t$ affects the function $f$.
