What happens to $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for functions over curves in $\mathbb{R^2}$? Suppose we have $f(x,y)=x+y$, and $\operatorname{domain}(f)=\{(x,y) \mid y=2x\}$. Now, what happens to $\dfrac{\partial f}{\partial x}$ at some point in the domain, say $(x_0,y_0)$? Will it be some real number, $0$ or undefined? According to me, this should be undefined. I think so because one can write $f(x,y)=3x$, and conclude $$\dfrac{\partial f}{\partial x}=3$$
or one may write $f(x,y)=\dfrac{3}{2}y$, and conclude
$$\dfrac{\partial f}{\partial x}=0$$
However, if the function is defined over a curve, then finding the partial derivatives is meaningless, as keeping one quantity fixed is not possible. So, shouldn't this mean that these functions don't have partial derivatives?
The question arose in my mind while thinking of finding $\dfrac{df}{dt}$, for parametrisation $$x=t,y=2t$$
If I use total derivatives,
$$\dfrac{df}{dt}=\dfrac{\partial f}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}$$
then I am confused regarding the partials. Shouldn't this operation be undefined for such functions?
 A: This is my first answer here.
In the first case, when you speak of $y=2x$, you define a particular domain, namely, the line $y-2x=0$.
So the partial derivatives would not make any sense for $f$.
On the other hand, you are considering a parameterized curve, which is not the same as being a domain. In the latter case, you are simply describing a curve, while in the former, you are considering a domain. 
For example, it's like asking, if domain of t = {1}, what is \frac {df}{dt}?
In the first case, $x$ and $y$ are $inputs$, and in the second, $outputs$.
A: I think part of the difficulty here is that the notation for partial
derivatives is confusing and sometimes misleading. 
Similar concerns are discussed in
Better Notation for Partial Derivatives.
In many contexts where you'd find a total derivative of some function $f$ expressed in terms of partial derivatives of $f$
(possibly all such contexts, but I don't trust myself to say so),
the function $f$ is really and truly a function of $n$ parameters, $n > 1,$ defined on $\mathbb R^n$ or on an $n$-dimensional subset of $\mathbb R^n.$
For convenience, the function may be defined by some equation with
$f(x,y)$ on the left-hand side; when that particular format is used (it is not always!) the partial derivative of $f$ with respect to its first parameter may be written $\frac{\partial f}{\partial x}$.
But it must be remembered that $x$ in this notation specifies only the position of a parameter within the list of parameters of $f,$
and not some variable named $x$ that may have been defined elsewhere.
Now you may have some differential equation involving the function $f$
and some actual variables named $x$ and $y,$ both dependent on $t,$
(that is, $x$ and $y$ are themselves functions of $t$)
with a condition such as $x(0) = y(0) = 0,$
and the solution of the equation with this condition may give us functions 
$x = x(t)$ and $y = y(t)$ that follow some curve, $y = h(x).$ 
Then for this particular solution of the equation, we have
$$\frac{d}{dt} f(x,y) =
\frac{\partial f}{\partial x}\frac{dx}{dt} 
+ \frac{\partial f}{\partial y}\frac{dy}{dt}. \tag1$$
But there is a problem with this notation, because the poor little letter $x$ is being forced to work two completely different jobs:
in one place, $\frac{\partial f}{\partial x},$
it signifies that you take the partial derivative of the two-parameter
function $f$ with respect to its first parameter;
but in the other two places, $x$ names a particular function of $t$
that is part of the solution of a particular differential
equation with a particular condition.
Note that in this application, the domain of $f$ is not the curve $y=h(x).$
Indeed, it would not make sense to restrict the domain of $f$ to that particular curve before using the differential equation to find out that the curve $y=h(x)$ is what we were looking for, because we need to define a domain for $f$ first in order to use the differential equation.
Also note that the same differential equation would produce a different solution for the condition $x(0)=y(0)=1$ (provided that it can be solved at all under that condition),
producing a different pair of functions $x(t)$ and $y(t)$
that would satisfy Equation $1$ above.
So I would contend that there is a problem with Equation $1,$
but the problem is a relatively minor problem in the notation,
namely, a confusing convention regarding how we notate the partial derivatives of a two-parameter function $f,$ not an error in the way we define the partial derivatives of the function $f.$

On the other hand, if we are in a situation where we already know at the very beginning that $x$ and $y$ must satisfy $y = h(x),$
then I think we have no business using partial derivatives.
I think instead we might define a function $g(t) = f(x(t),y(t)),$ and after that deal only with the single-parameter function $g$ instead of the two-parameter function $f.$
