Suppose $\mathbb{K}$ is a field of characteristic $0$. Let $S_n=\mathbb{K}[[x_1,\dots,x_n]]$ be the ring of formal power series in $n$ variables, $L_n$ be its fraction field and $W_n=\mathbb{K}[x_1,\dots,x_n,\partial_1,\dots,\partial_n]$ be the $n$-dimensional Weyl algebra, acting on $S_n$ and $L_n$ by multiplication and formal partial differentiation. An element $f \in L_n$ is called differentially integral if $M_f=\{ T(f)|T \in W_n \}$ is a finitely generated $P_n=\mathbb{K}[x_1,\dots,x_n]$-module. Let $R_n=\mathbb{K}(x_1,\dots,x_n)$, the fraction field of $P_n$ and denote the set of differentially integral elements by $DZ_n$.

Questions:

1. Does $DZ_n$ form a strict subring of $S_n$? It is an easy exercise to show that $DZ_n \cap R_n = P_n$ and $P_n \subsetneq DZ_n$. It is also known that $S_n \backslash DZ_n$ is non-empty as in this post.
2. Since $W_n$ acts on $L_n$ linearly, can we say that $DZ_n$ is to some extent 'graded' by $W_n$ via filtrations like $\{ T_1(f)=0 \} \subseteq \{ T_2T_1(f)=0 \} \subseteq \cdots$ if $DZ_n$ is an actual ring? When it is only allowed to use pure differential, the grading restricted to $P_n$ is precisely the multi-degree.
3. Change the requirement to be $M_f$ spanning a finite-dimensional vector space over $R_n$. Call these $f$ differentially algebraic and denote the set of them by $DA_n$. If $DZ_n$ is an actual ring, does $DA_n$ coincide with the field of fractions of $DZ_n$? This comes from the observation that $R_n \subsetneq DA_n$. However, there are also elements in $DA_n$, for example $f=\ln (1-x)$ when $\mathbb{K}=\mathbb{C},n=1$, that do not seem to come from any quotient in $DZ_n$, although a careful analyse is needed to verify this case.