Given a rotation matrix $A\in \text{SO}(2)$, there is a unique transformation $r_A:\mathbf{R}^6\to\mathbf{R}^6$ such that $$r_A(f(0),\frac{\partial f}{\partial x}(0),\dots,\frac{\partial^2 f}{\partial y^2}(0))=(f_A(0),\frac{\partial f_A}{\partial x}(0),\dots,\frac{\partial^2 f_A}{\partial y^2}(0))$$ for all smooth $f:\mathbf{R}^2\to\mathbf{R}$ where $f_A(x)$ denotes $f(Ax)$. For example, $f_A(0)=f(0)$, $[\frac{\partial f_A}{\partial x}(0),\frac{\partial f_A}{\partial y}(0)]=A^{-1}[\frac{\partial f}{\partial x}(0),\frac{\partial f}{\partial y}(0)]$ (I think), etc.
Which $g:\mathbf{R}^6\to\mathbf{R}$ are invariant under pre-composition with $r_A$?
One such example is $\frac{\partial f}{\partial x}(0)^2+\frac{\partial f}{\partial y}(0)^2$, i.e. the Laplacian of $f$ evaluated at the origin. Another one would be $\frac{\partial^2 f}{\partial x^2}(0)+\frac{\partial^2 f}{\partial y^2}(0)$, if I'm not mistaken.
Is there maybe an invariant basis $b_1,\dots,b_k:\mathbf{R}^6\to\mathbf{R}$ such that each invariant $g$ can be represented as $g^*:\mathbf{R}^k\to\mathbf{R}$ with $g(f(0),\frac{\partial f}{\partial x}(0),\frac{\partial f}{\partial y}(0),\dots,\frac{\partial^2 f}{\partial y^2}(0))$ being equal to $$g^*(b_1(f(0),\frac{\partial f}{\partial x}(0),\dots,\frac{\partial^2 f}{\partial y^2}(0)),\dots,b_k(f(0),\frac{\partial f}{\partial x}(0),\dots,\frac{\partial^2 f}{\partial y^2}(0)))?$$
NOT A COMPLETE ANSWER.
The following result is Theorem 8.61, p.275 in Folland "Real Analysis" (1984). It partially answers your question.
The result is: let $L=\sum_{\lvert \alpha\rvert\le m} a_\alpha(x)\partial^\alpha $ be a differential operator on $\mathbb R^d$ (not necessarily with constant coefficients). It holds that $$ L\circ T = T\circ L$$ for all rotations and translations $T$ of $\mathbb R^d$ if and only if $L=p(-\Delta)$ for some polynomial $p$. In words, the only differential operators that are invariant under rotations and translations are polynomials in the Laplacian.
This is not complete, because you require only rotation invariance and not translation invariance. However, I am rather sure that the proof can be adapted to yield a nice characterization also in your case, and this is the reason why I am posting this incomplete answer.