Here is another approach, which I believe I first learned from Toby Bartels. Suppose $X$ is an arbitrary differentiable manifold (think of the state space of some physical system), and define a *variable* (one might also say "observable") to be a smooth real-valued function on $X$. If $x:X\to \mathbb{R}$ is such a "variable", then its *differential* is, as usual in differential geometry, a smooth function ${\rm d}x:T X \to \mathbb{R}$ on the tangent bundle of $X$. We also have the tangent map $T x : T X \to T\mathbb{R} \cong \mathbb{R}\times\mathbb{R}$, with $T x = (x, {\rm d}x)$.

If $y:X\to \mathbb{R}$ is another such "variable", then it might be related to $x$ by an equation such as $y = x^2$ or $x^2 + y^2 = 4$. Being equalities of real-valued functions, these are *pointwise* equalities. If $y= x^2$, then we can say that "$y$ is a function of $x$" in the sense that there is a function $f:\mathbb{R}\to\mathbb{R}$ such that $y = f\circ x$, namely $f = \lambda u. u^2$ (see this question). In this case, the chain rule of differential geometry tells us that $T y:T X \to T \mathbb{R}$ is the composite $T X \xrightarrow{T x} T \mathbb{R} \xrightarrow{T f} T \mathbb{R}$. Since $T f (u,v) = (f(u), f'(u) \cdot v)$, this means that (in addition to $y = f\circ x$) we have ${\rm d}y = f'(x) \cdot {\rm d}x$. This is a simple pointwise equality of functions $T X \to \mathbb{R}$, so we can divide by ${\rm d} x$ (at least assuming it is never zero) to get $f'(x) = \frac{{\rm d}y}{{\rm d}x}$, or in this case $\frac{{\rm d}y}{{\rm d}x} = 2x$.

Similarly, if $x^2+y^2=4$, then $y$ is not a function of $x$ in this sense, but $x^2+y^2$ and $4$ are two smooth functions $X\to \mathbb{R}$, where the first is expressed as a composite $$X\xrightarrow{(x,y)} \mathbb{R}\times\mathbb{R} \xrightarrow{\lambda u v. u^2+v^2} \mathbb{R}.$$ Thus the chain rule of differential geometry again gives us $2 x \,{\rm d}x + 2 y \,{\rm d}y = 0$ as a pointwise equality of functions $T X \to \mathbb{R}$, so that we can solve it as usual in elementary calculus to get $\frac{{\rm d}y}{{\rm d}x} = -\frac{x}{y}$.

Calculusin a bygone century that it had a lot to say about variables. Don't recall exactly what. $\endgroup$2more comments