Let $k$ be an algebraically closed field and $V \subseteq k^n$ an affine variety corresponding to a prime ideal $P \subseteq k[t_1, \dots, t_n]$. For $x\in V$ let $O_x = \{p/q \mid p,q\in k[t_1, \dots, t_n]/P, q(x) \neq 0 \}$ and $m_x \subseteq O_x$ the maximal ideal of all $f\in O_x$ with $f(x) = 0$. Choose a basis $u_1, \dots, u_m$ of the vector space $m_x/m_x^2$. Following the definition in Basic Algebraic Geometry 1, Igor R. Shafarevich, p. 101 a Taylor series of $f\in O_x$ is a formal expression $F_0 + F_1 + \dots $ where $F_i$ is a homogeneous polynomial in $k[T_1, \dots, T_m]$ of degree $i$ such that
$f - S_l(u_1, \dots, u_m) \in m_x^{l+1}$ with $S_l = \sum_{i=0}^l F_i$.
If $x$ is non-singular, then the Taylor series is unique. Now fix $f = p/q$ with $p,q\in k[t_1, \dots, t_n]/P$. We may consider the Taylor series $F_{0,x}+F_{1,x}+\dots$ at different non-singular points $x\in V$ with $q(x) \neq 0$. I have the following question.
Can the coefficients in $F_{i,x}$ be expressed as $p'(x)/q'(x)$ for appropriate $p', q' \in k[t_1, \dots, t_n]/P$?
For example we always have $F_{0,x} = f(x)$ and the case $V = k$ is simply $F_{i,x} = \frac{1}{n!}f^{(n)}(x)T^i$. Note that the question is ill-posed, since the Taylor series depends on the choice of a basis of $m_x/m_x^2$ in each point $x\in V$. (For the case $V = k$ we chose $t-x \in m_x$ as a basis.) So we should additionally depand that we choose the basis appropriately to get the property in the question. If this is not possible globally, then maybe at least locally?