Solve without using logarithmic function

$4^{xy - (x + y)} = \cfrac{4^{xy}}{4^{x + 1}} = \cfrac{(4^y)^x}{4^x \cdot 4} = \cfrac{(12)^x}{4^x \cdot 3^x} = \cfrac{12^x}{12^x} = 1 = 4^0$

So $xy - (x + y) = 0$ , which rearranges to $\cfrac{xy}{x + 1} = \boxed{1}$ .

Note that $3^{xy} = 4^y = 12$ , and that $3^{x+1} = 3(4) = 12$ . Then:

$\begin{aligned} \dfrac{3^{xy}}{3^{x+1}} &= \dfrac{12}{12}\\ 3^{xy-(x+1)} &= 1\\ 3^{xy-(x+1)} &= 3^0\\ xy -(x+1) &= 0\\ xy &= x+1\\ \dfrac{xy}{x+1} &= \boxed{1} \end{aligned}$