If $z = f(x, y)$, then why are $\partial_x z$ and $\partial_y z$ functions of x and y also?

Question

Source: Stewart, James. Calculus: Early Transcendentals (6 edn 2007). p. 905.

Stewart doesn't explain the tree diagram beneath. Please explain intuitively and informally.

Answer 1

Simply put, the function is defined on a domain, so the derivatives, where they exist, should be defined on that domain as well.

Now, it may be that a function has partial derivatives that are not functions of $x$ and $y$, for instance: $$f(x,y) = x^2+y^2.$$

However, when we're talking about $\frac{\partial f}{\partial x}$, we want to ensure that we're talking about the same domain as $f$. So if $f : \mathbb{R}\times \mathbb{R} \to \mathbb{R}$, it could be somewhat awkward if we talk about features of the function that aren't also mapping from $\mathbb{R}\times\mathbb{R}$, as well. Now, with the example of the function above, it turns out that one of the co-ordinates in the domain doesn't matter, but in general we don't want to start from that assumption.