Just a random thought, for $A\subseteq\mathbf{R}^n$ with $0\in A$ we can define a derivative of a function $f:A\to\mathbf{R}$ at $0$ as a linear function $g:\mathbf{R}^n\to\mathbf{R}$ such that $\lim_{x\to 0}(g(x)-f(x))/\|x\|=0$ but this $g$ doesn't have to be unique. This is the case of course when e.g. $A$ is some proper linear subspace of $\mathbf{R}^n$, but for a less trivial example, consider $A:=\{x\in\mathbf{R}^2:|x_1|\le x_0^2\}$ and $f(x):=0$. Then $f$ itself is one derivative, but also $g(x):=x_1$, as $$\frac{|g(x)-f(x)|}{\|x\|}=\frac{|x_1|}{\sqrt{x_0^2+x_1^2}}\le\frac{x_0^2}{\sqrt{x_0^2}}=x_0$$ and $\lim_{x\to 0}x_0=0$.
So, my question, what property does $A$ need to satisfy such that all functions on $A$ have at most one derivative at $0$? My thought was something like there exists $\delta>0$ such that for $\epsilon>0$ there exists $P\in A^n \subseteq\mathbf{R}^{n\times n}$ with $\|P\|<\epsilon$ and $\det(P)\ge\delta\|P\|$ for some fixed matrix norm $\|.\|$. In other words, $A$ contains arbitrary small bases of $\mathbf{R}^n$ whose determinant is not arbitrary small in relation to the length of the vectors in that base.
Is this characterization correct and/or is there a simpler way to express this property?
[EDIT] Aphelli noted that $\det(P)\in O(\|P\|^n)$ so I guess it's better to say $\det(P)\ge\delta\|P\|^n$?
It really depends on how many "directions" we can approach $0$ via sequences in $A$. I will do this in a general setting.
To make this more formal, let $V$ and $W$ be completely arbitrary normed vector spaces, and $A \subseteq V$. We can define a unit vector $u$ to be a limiting direction (at $0$) for $A$ if there exists a sequence $(x_n)_{n=1}^\infty$ in $A \setminus \{0\}$ with $x_n \rightarrow 0$ as $n \rightarrow \infty$, such that:
$$u = \lim_{n \rightarrow \infty} \frac{x_n}{||x_n||_V}$$
Define $\text{Limit}(A)$ to be the linear span of all limiting directions of $A$. Then:
Edit: Someone requested a proof for the if and only if claim, so I am adding this below.
First assume that $0 \in A$ and $\text{Limit}(A) = V$. Let $f : A \rightarrow W$ be function which is differentiable at $0$, with derivatives $L$ and $M$. We will prove that $L = M$.
Let $u$ be a limiting direction for $A$, and let $(x_n)_{n=1}^\infty$ be a sequence as given in the definition of limiting direction. Then, by definition of derivative, we must have:
where $g$ and $h$ are two functions satisfying $g(n) \rightarrow 0$ and $h(n) \rightarrow 0$ as $n \rightarrow \infty$.
Then:
$$L(x_n) + ||x_n||_V g(n) = M(x_n) + ||x_n||_V h(n)$$
$$\frac{1}{||x_n||_V} L(x_n) + g(n) = \frac{1}{||x_n||_V} M(x_n) + h(n)$$
$$L(\frac{x_n}{||x_n||_V}) + g(n) = M(\frac{x_n}{||x_n||_V}) + h(n)$$
Then by taking the limit as $n \rightarrow \infty$ on both sides, we find $L(u) = M(u)$ (here I have used the fact that bounded linear functions are continuous).
Therefore $L$ and $M$ agree on every limiting direction of $A$, and hence agree on every linear combination of these vectors. But that is $\text{Limit}(A) = V$, so $L = M$.
For the other direction, note the claim is obvious if $W = \{0\}$. So assume that $W \not = \{0\}$.
If $\text{Limit}(A) \not = V$, then $V$ cannot be the trivial vector space, and let $v \in V$ be such that $v \not \in \text{Limit}(A)$ and $v \not = 0$. Let $w \in W$ with $w \not = 0$.
Now consider the zero function $f : A \rightarrow W$ given via $f(x) := 0$. Clearly the the zero map is a derivative for $f$ at $0$.
However we can construct a second distinct derivative to $f$ as follows:
Then $D$ is a second distinct derivative for $f$ at $0$.
(In fact given any function $g : A \rightarrow W$ with a derivative $E$ at $0$, then $E + \lambda D$ is also a derivative at $0$ for any scalar $\lambda$. So not only does there exist a function with a non-unique derivative, but every function either has no derivative at $0$ or uncountably many).