Is $f(x,y) = f(\mathbf{x})$ abuse of notation?
A scalar function $f(x,y)$ is often written as $f(\mathbf{x})$, where $\mathbf{x} = (x,y)$, but as far as I know, there is a difference between the scalar function inputs $(x,y)$ and the vector input $(x,y) = x\imath+y\jmath$. As I see it $f(\mathbf{x}) = f((x,y)) = f(x\imath+y\jmath) \neq f(x,y)$. Am I wrong, or is there a simple bijection between the two concepts?
Is it simply shorthand for $f': \mathbf{x} \mapsto f''(\imath\cdot\mathbf{x},\jmath\cdot\mathbf{x})$, s.t. $f'(\mathbf{x})= f''(x,y)$?
If it's of an relevance, I'm reading about scalar fields, and this definition came up:
$\displaystyle\dfrac{\partial f}{\partial x}(x,y) = \lim_{h\to 0} \dfrac{f(x+h, y)-f(x,y)}{h} \overset{\color{green}{?}}{=} \dfrac{f(\mathbf{x}+h\imath)-f(\mathbf{x})}{h} = \dfrac{\partial f}{\partial\imath}(\mathbf{x})$
While it looks nice I'm just curious if it's correct. However I don't see how $f$ can be differentiated with respect to both $\imath$ (a vector) and $x$ (a scalar), unless it's actually two different functions $f'$ and $f''$...
From page 32 of Halmos's book, Naive Set Theory:
$f(\mathbf{x}) = f((x, y))$ is more commonly written as $f(x, y)$ (where $\mathbf{x}=(x, y)$ is a vector in e.g. $\mathbb{R}^2$). The same thing happens with longer tuples: $f(x_1, \ldots, x_n)$. Pedantically speaking, $f((x, y))$ is indeed the (more) correct notation. I suppose the reason for omitting the extra parenthesis (besides convention) is that they are simply reduntant; removing them causes no ambiguity whatsoever.
I would not call this abuse of notation as we are just using two different notations for the same thing.
We are basically just switching back and forth between vectors $\vec x = xi+yj $ and the coordinates of the vectors $(x,y)$ with respect to a certain basis $(i,j)$. (Usually the basis is implicitly the cannonical one, or can be derived from the context.) As the coordinates of a vector with respect to a given basis are unique, this is does not allow any ambiguity.
Your example is indeed correct, and it obviously depends on how whether you define first a vectorial derivative or first the partial derivative. In any case, the derivative in the direction of a basis vector is equal to the derivative with respect to the corresponding coordinate.
$\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Reals}{\mathbf{R}}$Literally, you're correct: There is a meaningful distinction between
A real-valued function $f$ accepting two real numbers as input;
A real-valued function $g$ accepting an ordered pair of real numbers as input.
Just as you say, the literal notations would be $f(x, y)$ and $g(\Vec{x}) = g\bigl((x, y)\bigr)$, respectively. Strongly-typed computer languages such as C++ enforce this distinction.
Even in pure mathematics, if $f$ is a real-valued function accepting two real numbers as input, you can guiltlessly form, e.g., $$ f\bigl(x, f(y, z)\bigr),\qquad f\bigl(f(x, y), z\bigr). $$ By contrast, if $g$ accepts ordered pairs of reals as input and returns real values, it would feel at least incrementally wrong to write $$ g\bigl(x, g(\Vec{y})\bigr),\qquad g\bigl(g(\Vec{x}), z\bigr). $$ (Note to self: In mathematics, one wrong doesn't make a write.)
Luckily, we're saved from serious ambiguity by the natural identification of ordered pairs of real numbers (two real things, $\Reals \times \Reals$) with ordered pairs of real numbers (one vector thing, $\Reals^{2}$). Practically, this identification allows us to drop the second set of parentheses, and write $g(x, y)$ instead of $g\bigl((x, y)\bigr)$ as convenience dictates.
A viewpoint that hasn't been mentioned yet:
We can take $f\mathbf{x}$ and $f\langle x,y\rangle$ as the "correct" notations, and then stipulate that we can include round brackets wherever we want, as disambiguators. Under these conventions, the notations $$f(\mathbf{x}),\quad f\langle x,y\rangle$$ are perfectly acceptable.
This also gives us notation for "global" elements; if $x \in X$, then the corresponding function $1 \rightarrow X$ can be denoted $\langle x \rangle,$ since by definition, this is an element of $X^1$.