Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be bicategories. We say a 'pseudofunctors of 2-variables' consists of two families of pseufunctors
Fix $A\in obj\mathcal{A}$, we have a pseudofunctor $f(A,-): \mathcal{B}\to \mathcal{C}$.
Fix $B\in obj\mathcal{B}$, we have a psudofuncotr $f(-,B): \mathcal{A}\to \mathcal{C}$.
$f(A,-)(B)=f(-,B)(A):=f(A,B)$ for all $A\in obj\mathcal{A}, B\in\mathcal{B}$.
for all $\alpha: A\to A',\beta:B\to B'$, an invertible 2-cell $f(\alpha,\beta)$ between $$ f(A,B)\xrightarrow{f(A, \beta)}f(A,B')\xrightarrow{f(\alpha,B')} f(A',B'),\\ f(F,B)\xrightarrow{f(\alpha, B)}f(A',B)\xrightarrow{f(A',\beta)}f(A',B'). $$
These two families and $f(\alpha,\beta)$ satisfy coherence conditions similar to that of quasi-functor of two variables introduced by Gray ("Adjointness for 2-Categories", page 56), but droping the first requirement on identities, which I shall not spell out here; see Gordon-Power-Street in "Coherence for tricategory", where it is called cubical functors (so remove the line "which is an identity when either $\alpha$ or $\beta$ is an identity).
Notice that if $\mathcal{A}=\mathcal{B}=*$, the trivial bicategory with one objects, one 1-cell, one 2-cell, then the above data should be equivalent to two monads in $\mathcal{C}$ on the same object satisfying a distributive law.
Question:
Does there exist a funny tensor product $\mathcal{A}\otimes \mathcal{B}$ such that the above data is the same as a single pseudofunctor $\mathcal{A}\otimes \mathcal{B}\to \mathcal{C}$? (Gray tensor product of bicategories?)