Let $R$ be an integral domain of characteristic zero and let $A=A(x,y) \in R[x,y]$ with $\deg(A) \geq 1$ (total degree).
Assume that there exist $w=w(x,y),C=C(x,y) \in R[x,y]$ such that $w_x=CA_x$ and $w_y=CA_y$. The subscripts denote partial derivatives, of course.
Can one find the exact form of such $w$ and $C$?
Notice that a possible solution is as follows: $w=\sum_{i=0}^{m}\lambda_iA^i$, for some $m \in \mathbb{N}$, $\lambda_i \in R$, $1 \leq i \leq m$ (in short, $w=f(A)$, for some $f \in R[T]$), and $C=\sum_{i=1}^{m}i\lambda_iA^{i-1}=f'(A)$. Indeed, $w_x=f'(A)A_x=CA_x$ and $w_y=f'(A)A_y=CA_y$.
Is it true that a solution is necessarily of the form $w=f(A)$, $C=f'(A)$?
Observations:
(i) $w_x=CA_x$ implies that $w_{xy}=C_yA_x+CA_{xy}$, so $w_{xy}-CA_{xy}=C_yA_x$.
(ii) $w_y=CA_y$ implies that $w_{yx}=C_xA_y+CA_{yx}$, so $w_{yx}-CA_{yx}=C_xA_y$.
Since $w_{xy}=w_{yx}$ and $A_{xy}=A_{yx}$, we get that $C_yA_x=C_xA_y$.
Also observe that $C_yA_x=C_xA_y$ implies that $C_yw_x=C_y(CA_x)=C(C_yA_x)=C(C_xA_y)=C_x(CA_y)=C_xw_y$.
Perhaps knowing that $C_yA_x=C_xA_y$ and $C_yw_x=C_xw_y$ may help solve my question.
Remark: I am especially interested in a non-normal integral domain $R$ of characteristic zero such as $R=\mathbb{C}[t^2,t^3]$ or $R=\mathbb{Z}[\sqrt{5}]$, though I am not sure if this should affect the answer.
Any hints and comments are welcome!