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Background: Many textbooks give an example due to Peano of the function $f(x, y) = xy(x^2 - y^2)(x^2 + y^2)^{-1}$ that has mixed partial derivatives at $0$ that are not symmetric. One might wonder how arbitrary mixed partial derivatives can be at a point. For what choices of an appropriate number of real numbers is there a real-valued function that takes on those values for its partial derivatives at a point?

Notation:

Let $d\in \mathbb{N}$ and let $[d] = \{1,\ldots,d\}$. Let $\mathcal{I}$ be the set of all $n$-tuples with elements in $[d]$ (that is, $\mathcal{I} = \{(i_{1},\ldots, i_{n}): i_{1},\ldots, i_{n}\in [d]\}$). Let $h:\mathcal{I}\rightarrow \mathbb{R}$ . For which functions $h$ is there a function $f:\mathbb{R}^{d}\rightarrow \mathbb{R}$ so that $$ \frac{\partial}{\partial x_{i_{n}}}\cdots \frac{\partial}{\partial x_{i_{1}}}f = h((i_{1},\ldots, i_{n})) $$ at $x = 0\in \mathbb{R}^{d}$

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Unless I'm misunderstanding your question, I think you're asking something along the lines of whether there exists a function of $d$ variables for which there exists a point (or a dense set of points, or some other "large" number of points) such that for each positive integer $n$ we have all "differently iterated" $n$'th order partial derivatives existing and different from each other at that point. And if not, then what are some constraints on values of $d$ and $n$ and permutations of $n$'th order partial derivatives that prevent such an example from being possible.

At this time I don't know of any reasonably nontrivial examples, but surely some extreme pathological examples have been published, and if not, this would be an excellent research topic. However, for now I'd like to use your question to rescue from oblivion (because not googleable) some information I gave in comments to this Mathematics Educators answer.

The function

$$f(x,y) = \exp\left(-\frac{x^2}{y^2} - \frac{y^2}{x^2}\right)\;\; \text{for} \;\; xy \neq 0, \;\; \text{with} \;\; f(x,0)=f(0,y)=0,$$

is such that all mixed partials exist and are independent of differentiation order, but $f(x,y)$ is not continuous at $(x,y)=(0,0).$ For example, we have

$$\frac{{\partial}^8f}{\partial x^2 \, \partial y^3 \, \partial x \, \partial y^2} \; = \; \frac{{\partial}^8f}{\partial x \, \partial y^2 \, \partial x \, \partial y^3 \, \partial x} \; = \; \frac{{\partial}^8f}{\partial x^3 \, \partial y^5} \; = \; \frac{{\partial}^8f}{\partial y^5 \, \partial x^3}$$

This example is given in solution to Problem 4876 in American Mathematical Monthly 67 #8 (October 1960), pp. 813−814 (ASCII version in this 21 December 2006 sci.math post).

Regarding the history of this topic, see the StackExchange History of Science and Mathematics question Who first proved that $f_{xy} = f_{yx}$? and the following paper:

Thomas James Higgins (1911−1998), A note on the history of mixed partial derivatives, Scripta Mathematica 7 (1940), pp. 59−62.

A scanned copy of the paper by Higgins is not available online, but I posted an ASCII version of his paper in this 9 June 2007 sci.math post.

Two papers probably worth looking at for this topic, especially their references, are:

Ettore Minguzzi, The equality of mixed partial derivatives under weak differentiability conditions, Real Analysis Exchange 40 #1 (2014−2015), pp. 81−97.

Marek Jarnicki and Peter Pflug, Directional regularity vs. joint regularity, Notices of the American Mathematical Society 58 #7 (August 2011), pp. 896−904.

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