Symmetry behind Taylor expansion We consider the function $f:\mathbb R^2 \to \mathbb C.$
$$f(x,y) = 2i \sum_{k=0}^2 \zeta^k \sin\left( \frac{x\bar \zeta^k-y\zeta^k}{2i} \right)$$
where $\zeta= e^{2\pi i/3}$ is the third root of unity
I computed using Mathematica the Taylor expansion at $x=y=0$ up to order $8$
$f(x,y) = \left(-\frac{y^5}{640}\right)+x
\left(3+\frac{y^6}{15360}\right)+x^2 \left(-\frac{3
y}{8}-\frac{y^7}{860160}\right)+x^3
\left(\frac{y^2}{64}+\frac{y^8}{82575360}\right)+x^4
\left(-\frac{y^3}{3072}\right)+x^5
\left(\frac{y^4}{245760}\right)+x^6
\left(-\frac{y^5}{29491200}\right)+x^7
\left(\frac{1}{107520}+\frac{y^6}{4954521600}\right)+x^8
\left(-\frac{y}{3440640}-\frac{y^7}{1109812838400}\right)+O\left(x^9+y^9
\right).$$
Then since the function is anti-symmetric by switching both $x,y$, we only get products of odd order in $x^iy^j$, i.e. $i+j$ is odd.
However, there is another feature that is more subtle for me to explain. For a given power in $x$ we have some $y^i$ and then a term $y^{i+6}$ appearing. Why do we have these jumps of order $6$ in the expansion?
Please let me know if you have any questions.
 A: Overview. Your problem is connected to the general problem of constructing analytic functions that possess prescribed symmetries. One trick for creating such functions is to average "any" analytic function over the action of all elements of the  desired symmetry group.  This trick compels the Taylor coefficients  of the symmetrized function to have some special symmetrical patterns.
Details. The crux of the matter is that the derived function $\partial f/\partial x$ is loosely speaking a symmetrization of $h(x,y)=\cos(\frac{x-y}{2i})$ under the action of a finite group generated by powers of $\zeta$.
More precisely you can verify the identity
(1) $f_x= g(x,y)=\sum_{k=0}^2 h( \zeta^{-k} x, \zeta^k y)$
Consider the  implications of the symmetry
(2) $h(-x,-y)=h(x,y)$
on the Taylor expansion
(3) $h(x,y)= \sum_{m,n\geq 0} c_{m,n} x^m y^n$
Deduce that $c_{m,n} =(-1)^{n+m} c_{m,n}$.
That is
(4)  $c_{m,n}=0$ unless $n$ and $m$ have the same parity (either both even, or both odd).
Substitute  (3) into (1) to deduce that
$g(x,y)= \sum_{k=0}^2 h(  \zeta^{-k} x, \zeta^k y)$ satisfies
(5) $g(x,y)= \sum_{m,n\geq 0}  c_{m,n}  x^m y^m (\sum_{k=0}^2 \zeta^{k(n-m)})$.
The rightmost sum is a three-term geometric series consisting of three consecutive powers of  $\zeta^{n-m}$.
That geometric series vanishes unless $n-m \equiv 0 $(mod 3).
Thus the non-vanishing terms in the Taylor expansion of $g(x,y)$ consist only  of those terms for which
(6) $n \equiv m$ (mod 3).
But as noted in (4),  the additional requirement that $n$ and $m$ must have the same parity  forces the condition that $c_{m,n}=0$ unless $n\equiv m $(mod 6).
Finally, this pattern satisfied by $f_x$ naturally leads to the pattern that you noted about $f$ itself. (Just integrate the series for $f_x$ term-wise to obtain $f$.)
