Given equation $x y+4 x-3 y-12=0$ can be written as
$$
\begin{aligned}
x(y+4)-3(y+4) & =0 \\
\Rightarrow \quad(y+4)(x-3) & =0
\end{aligned}
$$
Also, the equation $x y-3 x+4 y-12=0$ can be written as
$$
\begin{aligned}
x(y-3)+4(y-3) & =0 \\
\Rightarrow \quad(y-3)(x+4) & =0
\end{aligned}
$$

Hence, equations of the sides of the square are
$$
x=3, x=-4, y=3, y=-4
$$
So, the coordinates of the vertex are $A(-4,-4)$, $B(3,-4), C(3,3)$ and $D(-4,3)$.


$$
\begin{aligned}
& \therefore \text { Equation of } A C: y+4=\frac{3+4}{3+4}(x+4) \\
& \Rightarrow \quad y+4=x+4 \Rightarrow y=x \\
& \Rightarrow \quad x-y=0 \\
& \therefore \quad \text { Equation } B D: y+4=\frac{3+4}{-4-3}(x-3) \\
& \Rightarrow \quad y+4=-(x-3) \\
& \Rightarrow \quad y+4=-x+3 \\
& \Rightarrow \quad x+y+1=0 \\
&
\end{aligned}
$$
Hence, combined equation of diagonals is
$$
\begin{array}{rlrl}
& & (x-y)(x+y+1) & =0 \\
\Rightarrow & x^2+x y+x-x y-y^2-y & =0 \\
\Rightarrow & & x^2-y^2+x-y & =0
\end{array}
$$