Are independent variables really independent?

I am studying partial differential equations, in which the class notes includes the following statement:

$$x^2 + y^2 + (z-c)^2 = a^2$$
Here z is the dependent variable; x,y are independent variables; and a,c are arbitary constants.

Mathematically, all the variables depend on every other variable, so there shouldn't be any difference between x,y or z. Is this a subjective distinction? Aren't variables objectively all dependent on each other?

It means that $z$ is a function of $x$ and $y$, so the full writing would be

$$x^2+y^2+(z(x,y)-c)^2=a^2$$

It's an unfortunately common abuse of notation that confuses the notion of "dependent" with one of the approaches for dealing with dependence.

The dependent triple $\{x,y,z\}$ has two degrees of freedom meaning that, for the most part, you can pick two variables $s$ and $t$ so that $\{s,t\}$ are an independent pair, and write all of $x,y,z$ as functions of $s$ and $t$.

(e.g. you could pick $s,t$ so that $s=x$ and $t=y$)

The abuse of notation, then, is to call $s,t$ the "independent variables" and $x,y,z$ the "dependent variables". (or, if you picked $s=x$ and $t=y$ to call $x,y$ the "independent variables" and $z$ the "dependent variable")

The intention is to then not work with the variables $x,y,z$, but instead the functions that express them in terms of $s,t$.

There is another unfortunate abuse of notation that further confuses the issue — the function expressing, for example, $z$ as a function of $s$ and $t$ is often named $z$ as well.

When god created the world ${\mathbb R}^3$ the three variables $x$, $y$, $z$ used to address the points $p\in{\mathbb R}^3$ were truly independent.

Now an equation $$x^2+y^2+(z-c)^2=a^2\tag{1}$$ is given, whereby for simplicity we assume $a>0$. Such an equation does not define any function per se, but a solution set $$S:=\bigl\{(x,y,z)\,\bigm|\,x^2+y^2+(z-c)^2=a^2\bigr\}\ .$$ In the case at hand this set is a $2$-sphere of radius $a$ with center $(0,0,c)$.

Nevertheless, an equation like $(1)$ does implicitly define a suitably chosen of the three variables as functions of the other two, albeit only locally. The implicit function theorem guarantees this (under "technical assumptions") even if you are not able to solve $(1)$ explicitly for the chosen variable, say $z$. To be precise:

If $f\in C^1$, and $\nabla f(x_0,y_0,z_0)\ne(0,0,0)$ at some point $(x_0,y_0,z_0)\in S$, say $f_z(x_0,y_0,z_0)\ne0$, then there is a rectangular box (a "local window")$$W=[x_0-h,x_0+h]\times[y_0-h',y_0+h']\times[z_0-h'',z_0+h'']$$ with center $(x_0,y_0,z_0)$, and a $C^1$-function $$\phi:\>[x_0-h,x_0+h]\times[y_0-h',y_0+h']\to[z_0-h'',z_0+h'']$$ such that $$S\cap W=\bigl\{(x,y,z)\,\bigm|\,|x-x_0|\leq h,\ |y-y_0|\leq h', \ z=\phi(x,y)\bigr\}\ .$$ This means that within the box $W$ the given equation $f(x,y,z)=0$ defines $z$ "implicitly" as a function $z=\phi(x,y)$.