3 v/s 4

It is given that:

$\begin{cases} (3x)^{\log 3} = (4y)^{\log 4} & ...(1) \\ 4^{\log x} = 3^{\log y} & ...(2) \end{cases}$

Taking logarithm on both sides of both the equations.

$\begin{aligned}(1): \quad \quad \quad \quad \ \log 3 (\log 3 + \log x) & = \log 4 (\log 4 + \log y) \\ \log^2 3 + \log 3 \log x & = \log^2 4 + \log 4 \log y \quad ... (4) \\ (2): \quad \quad \quad \quad \quad \quad \quad \quad \color{#3D99F6}{\log 4 \log x} & = \color{#3D99F6}{\log 3 \log y} \quad \quad \quad \quad \ \ ...(5) \\ (4) \times \log 3: \quad \log^3 3 + \log^2 3 \log x & = \log 3 \log^2 4 + \log 4 \color{#3D99F6}{\log 3 \log y} \\ \log^3 3 + \log^2 3 \log x & = \log 3 \log^2 4 + \log 4 \color{#3D99F6}{\log 4 \log x} \\ \log^2 3 \log x - \log^3 4 \log x & = \log 3 \log^2 4 - \log^3 3 \\ (\log^2 3 - \log^3 4 )\log x & = \log 3 (\log^2 4 - \log^2 3) \\ \implies \log x & = - \log 3 \\ \implies x & = 3^{-1} = \frac{1}{3} \\ (5): \quad \quad \quad \quad \quad \quad \quad \quad \color{#3D99F6}{- \log 4 \log 3} & = \color{#3D99F6}{\log 3 \log y} \\ \implies y & = 4^{-1} = \frac{1}{4} \end{aligned}$

Therefore, $\implies \dfrac{x+y}{x-y} = \dfrac{\frac{1}{3}+\frac{1}{4}}{\frac{1}{3}-\frac{1}{4}} = \boxed{7}$