This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?.

For any field $k$, the field $k(x)$ of rational functions in one variable has an explicit $k$-basis given by partial fractions: the set $$B(x,k)=\left\{x^i:i\geq 0\right\}\cup \left\{\frac{x^i}{m(x)^j}:\mbox{$m(x)$ monic irreducible, } 0<j, 0\leq i<\operatorname{deg}(m)\right\}$$ is a basis.

By induction, $k(x_1,\dots,x_n)$ has a $k$-basis for any $n$, without needing to assume the axiom of choice. The obvious way of doing this gives a basis that depends on the choice of an order for the variables.

In ZF set theory without choice, must $k(X)$ have a $k$-basis for an arbitrary set $X$?

The existence of such a basis certainly doesn't need the full strength of AC. The statement that every set has a total order is known to be independent of ZF but not to imply AC (see for example the various references in Are all sets totally ordered ?). If $X$ has a total order, and for $y\in X$ we let $$X_{<y}=\left\{x\in X: x<y\right\},$$ then the set of finite products $t_1t_2\dots t_n$, where $$1\neq t_i\in B\left(y_i,k(X_{<y_i})\right)$$ for some $y_1<\dots < y_n$, is a $k$-basis of $k(X)$.