Let $f:D\to\mathbb R$ be the following rational function $$ f(x_1,\ldots,x_n)=\frac{P(x_1,\ldots,x_n)}{Q(x_1,\ldots,x_n)}, $$ where here $P,Q$ are polynomials and $$ D=\{(x_1,\ldots,x_n):Q(x_1,\ldots,x_n)\neq0\}. $$ My question: I want to prove that if $\frac{\partial f}{\partial x_1}=0$ for all $(x_1,\ldots,x_n)\in D$, then $f$ is independent of $x_1$.
My attempt: I saw in the posts 1 and in 2 that this is not true for $f$ in general, but if $D$ is convex, then it is true. Can I assume that $D$ is convex in the case of rational functions? If not, is there another way to prove that claim?