Suppose that $f_1(x),f_2(x)$, $g_1(y),g_2(y)$ are non-constant rational functions and let
$$
F(x,y)=(f_1(x)+g_1(y))(f_2(x)+g_2(y)).
$$
I want to prove that $F$ must depend upon $x$, that is, that $\frac{\partial F}{\partial x}\not\equiv0$.
My attempt: Suppose otherwise that $F(x,y)=\phi(y)$, where $\phi$ is some rational function. Note that $$ \begin{split} (1)\quad & \frac{\partial f_1}{\partial x}(f_2+g_2)+\frac{\partial f_2}{\partial x}(f_1+g_1)\equiv0\\\\ (2)\quad & \frac{\partial g_1}{\partial y}(f_2+g_2)+\frac{\partial g_2}{\partial y}(f_1+g_1)\equiv\phi'(y)\\ \end{split} $$ that is $$ \underbrace{\begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_2}{\partial x} \\\\ \frac{\partial g_1}{\partial y} & \frac{\partial g_2}{\partial y} \\ \end{pmatrix}}_{M} \begin{pmatrix} f_2+g_2 \\\\ f_1+g_1 \\ \end{pmatrix}= \begin{pmatrix} 0 \\\\ \phi'(y) \\ \end{pmatrix}. $$ Furthermore, by (1) $$ (3)\qquad\frac{\partial f_1}{\partial x}\frac{\partial g_2}{\partial y}+\frac{\partial f_2}{\partial x}\frac{\partial g_1}{\partial y}\equiv0. $$ I think that it suffices to prove that $det M\equiv0$, since then, along with (3) we will deduce that $\frac{\partial f_1}{\partial x}\frac{\partial g_2}{\partial y}\not\equiv0$, a contradiction since $f_1,g_2$ are non-constant.
If your expression is only a function of $y$ then so must be $$f_1(x)f_2(x)+g_2(y)f_1(x)+g_1(y)f_2(x)$$
Hence $\frac{\partial}{\partial x}\Big(f_1(x)f_2(x)+g_2(y)f_1(x)+g_1(y)f_2(x)\Big)=0$ which unveils an important relationship: $$-\frac{f_2(x)+g_2(y)}{f_1(x)+g_1(y)}=\frac{f_2'(x)}{f_1'(x)}$$ Clearly $\frac{\partial}{\partial y}\Big(\frac{f_2(x)+g_2(y)}{f_1(x)+g_1(y)}\Big)=0$ and so $$\frac{g_2'(y)}{g_1'(y)}=\frac{f_2(x)+g_2(y)}{f_1(x)+g_1(y)}$$ But then $\frac{\partial}{\partial x}\Big(\frac{f_2(x)+g_2(y)}{f_1(x)+g_1(y)}\Big)=0$ which means $$\vec{\nabla}\Big(\frac{f_2(x)+g_2(y)}{f_1(x)+g_1(y)}\Big)=\vec{0}$$ It follows that $$(x,y)\mapsto \frac{f_2(x)+g_2(y)}{f_1(x)+g_1(y)}$$ is a constant function. Can you take the wheel from here?