Suppose that $f_1(x,z),f_2(y)$, $g_1(x,y),g_2(z)$ are non-constant rational functions and let
$$
F(x,y,z)=\big[\overbrace{f_1(x,z)+f_2(y)}^{f}\big]\big[\overbrace{g_1(x,y)+g_2(z)}^{g}\big].
$$
I want to prove that $F$ must depend upon $x$, that is, that $\frac{\partial F}{\partial x}\not\equiv0$.
Comment: By $f_1(x,z)$ we mean that $f_1$ is depend upon both $x$ and $z$. The same for $g_1(x,y)$, $f_2(y)$ and $g_2(z)$.
My attempt: Suppose otherwise that $F(x,y)=\phi(y,z)$, where $\phi$ is some rational function. Note that $$ \frac{\partial F}{\partial x}= f\frac{\partial g_1}{\partial x}+g\frac{\partial f_1}{\partial x}\equiv0 $$ $$ \frac{\partial^2F}{\partial x\partial y}= \frac{\partial f_2}{\partial y}\frac{\partial g_1}{\partial x}+ f\frac{\partial^2 g_1}{\partial x\partial y}+ \frac{\partial f_1}{\partial x}\frac{\partial g_1}{\partial y} \equiv0 $$ and $$ \frac{\partial^3F}{\partial x\partial y\partial z}= \frac{\partial f_1}{\partial z}\frac{\partial^2g_1}{\partial x\partial y}+ \frac{\partial^2f_1}{\partial x\partial z}\frac{\partial g_1}{\partial y} \equiv0. $$ Observe that the last equality is equivalent to $$ \frac{\partial}{\partial x}\left[ \frac{\partial f_1}{\partial z}\cdot \frac{\partial g_1}{\partial y} \right]\equiv0. $$ Therefore $\frac{\partial f_1}{\partial z}\cdot \frac{\partial g_1}{\partial y}$ does not depend upon $x$. How to continue from here?