Suppose we are doing two-variable calculus in $x,y$. Let $t=x+y$, and I would like to know how to calculate $\partial/\partial t$.
We know that $\partial t/\partial x=\partial t/\partial y = 1$. Using multivariable chain rule, $${\partial f\over \partial t} = {\partial f\over \partial x}{\partial x\over \partial t}+{\partial f\over \partial y}{\partial y\over \partial t} = {\partial f\over \partial x}+{\partial f\over \partial y}\tag{$*$}$$ However $${\partial (x^2-y^2)\over \partial t}={\partial t(x-y)\over \partial t}=x-y \tag{$**$}$$ while using ($*$) we would get $2x-2y$. So clearly something is wrong here...
My question is, what would be the correct way to write $\partial/\partial t$ in terms of $\partial_x$ and $\partial_y$?
This is a community answer, expanding the discussion from the comments. Please feel free to improve it to your liking!
The very first thing to remember is that a partial derivative is well-defined only after one fixes a coordinate system, because partial derivatives are coefficients appearing when expanding the differential form $\mathrm{d}f$ in a coordinate system:
$$ \mathrm{d}f = \sum_{k=1}^{n} \frac{\partial f}{\partial x_i} \mathrm{d}x_i $$
This tells that the partial derivatives are "linearly dependent" in some sense, and so, each of the partial derivatives does depend on the choice of the coordinate system. Let us demonstrate this using OP's example.
Example 1. Let $f = x^2 - y^2$, and choose the coordinate system $(t, s) = (x + y, x - y)$. Then
$$ \frac{\partial f}{\partial t} = \frac{\partial}{\partial t}(st) = s = x - y. $$
Example 2. Still let $f = x^2 - y^2$, but now choose the coordinate system $(t, x) = (x + y, x)$. Then
$$ \frac{\partial f}{\partial t} = \frac{\partial}{\partial t} t(2x-t) = 2x - 2t = -2y. $$