Confusing Notation of Partial Derivative Suppose that $z=f(x,y)$ is given. where $f(x,y)=x^3+y^3$.
I am a bit confused about the notation of $z_x$ and $f_x$.
A lot of people use the below notation for $z_x$.
$$z_x=3x^2$$
But what if $y$ is related to $x$ in some equation. For example
$$y=x.$$
Then
$${\partial z\over\partial x}$$
should also consider variable $y$ since $y$ can be represented by $x$.
But in a similar situation, I saw  in a book
$${\partial f\over\partial x }=3x^2$$
which doesn't consider y as a variable.
Summarizing, given that
$z=f(x,y)$, $f(x,y)=x^3+y^3$, $y=x$ what is $z_x$ and $f_x$?
 A: Let's suppose that $y=y(x)$. Then the derivative of $f(x,y)$ with respect to variable $x$ is $$\frac{d f(x,y(x))}{dx}=3x^2+3y^2(x)\frac{dy(x)}{dx}$$
The partial derivative is just a notation, based on the chain rule. It is the derivative when you consider all other variables independent of $x$. You can see now that you can write $$\frac{d f(x,y(x))}{dx}=3x^2\frac{dx}{dx}+3y^2(x)\frac{dy(x)}{dx}=\frac{\partial f(x,y)}{\partial x}\frac{dx}{dx}+\frac{\partial f(x,y)}{\partial y}\frac{dy}{dx}$$
A: I would say that $f_x = 3x^2$ and $f_y = 3y^2$.
If it's written $z=x^3+y^3$ and the notation $z_x$ is used then I would accept $z_x=3x^2$ even if it's also said that $y=x.$
For the total derivative the notation $z' = \frac{dz}{dx}$ would be more appropriate:
$$z' = z_x x' + z_y y' = 3x^2 \cdot 1 + \left.3y^2\right|_{y=x} \cdot 1 = 6x^2.$$
A: This is where it helps to introduce the notation $\left(\frac{\partial z}{\partial x}\right)_y$ to assume $y$ doesn't vary as $z$ varies due solely to $x$ varying. (I've not seen this abbreviated as $(z_x)_y$, but that's presumably OK< especially if you explain its meaning.) So$$z=x^3+y^3\implies \left(\frac{\partial z}{\partial x}\right)_y=3x^2,\,\left(\frac{\partial z}{\partial y}\right)_x=3y^2$$is true even if $y$ is a function of x. These partial derivatives may not seem useful in that case, but allow me to show they are. The multivariate chain rule is$$dz=\left(\frac{\partial z}{\partial x}\right)_ydx+\left(\frac{\partial z}{\partial y}\right)_xdy,$$which in the special case $y=f(x)$ gives$$dy=f^\prime(x)dx\implies dz=\left[\left(\frac{\partial z}{\partial x}\right)_y+\left(\frac{\partial z}{\partial y}\right)_xf^\prime(x)\right]dx.$$This leads naturally to @Andrei's calculation.
