Suppose $U$ is an open subset of $\mathbb{R}^n$, and $f:U\to \mathbb{R}$. When $f$ is $C^2$ we know that the mixed partial derivatives are symmetric, i.e. $\partial_i\partial_jf= \partial_j\partial_if.$ But as it is famous the continuity of the 2nd order partial derivatives is not necessary for this to happen. For example if $\partial_if$, $\partial_jf$ exist on $U$ and they are both differentiable (in the sense of Fréchet) at some point $a\in U$ then $$\partial_i\partial_jf(a)= \partial_j\partial_if(a).$$
Now for the 3rd order partial derivatives we can obtain the symmetry if we assume that the 1st order partial derivatives of $f$ are differentiable on $U$ and its 2nd order partial derivatives are differentiable at $a$. Let me explain the proof for the particular case $$\partial_3\partial_2\partial_1f(a)= \partial_2\partial_1 \partial_3f(a).\tag{$\star$}$$ First as $\partial_1 f$ has 1st order partial derivatives in $U$ and they are differentiable at $a$ we have $$\partial_3\partial_2\partial_1f(a)= \partial_2\partial_3 \partial_1f(a).\tag{1}$$ Then since the 1st order partial derivatives of $f$ are differentiable in $U$ we have $\partial_3\partial_1f(x)= \partial_1\partial_3 f(x)$ for all $x\in U$. Hence we can differentiate to obtain $$\partial_2\partial_3\partial_1f(a)= \partial_2\partial_1 \partial_3f(a).\tag{2}$$ By combining (1) and (2) we get ($\star$).
As you can see the full force of differentiability of the 1st order partial derivatives of $f$ on all of $U$ is only used for the equality of the 3rd order partial derivatives appeared in (2). So my question is
Question: Can we prove the symmetry of 3rd order mixed partial derivatives of $f$ at $a$ by merely assuming that the 1st and 2nd order partial derivatives of $f$ exist on $U$ and they are all differentiable at $a$? If not, can you provide a counterexample? Finally, if the answer is positive, can we generalize it to higher order mixed partial derivatives?
