Simplify:

$\dfrac{\Big(x+\dfrac{1}{y}\Big)^a\Big(x-\dfrac{1}{y}\Big)^b}{\Big(y+\dfrac{1}{x}\Big)^a\Big(y-\dfrac{1}{x}\Big)^b}$

$\dfrac{\Big(x+\dfrac{1}{y}\Big)^a\Big(x-\dfrac{1}{y}\Big)^b}{\Big(y+\dfrac{1}{x}\Big)^a\Big(y-\dfrac{1}{x}\Big)^b}\\[1em] = \dfrac{\Big(\dfrac{xy + 1}{y}\Big)^a\Big(\dfrac{xy-1}{y}\Big)^b}{\Big(\dfrac{xy+1}{x}\Big)^a\Big(\dfrac{xy-1}{x}\Big)^b}\\[1em] = \dfrac{\Big(\dfrac{1}{y}\Big)^a \times \Big(\dfrac{1}{y}\Big)^b (xy + 1)^a \times (xy - 1)^b}{\Big(\dfrac{1}{x}\Big)^a \times \Big(\dfrac{1}{x}\Big)^b (xy + 1)^a \times (xy - 1)^b}\\[1em] = \dfrac{\Big(\dfrac{1}{y}\Big)^{a + b}}{\Big(\dfrac{1}{x}\Big)^{a + b}}\\[1em] = \Big(\dfrac{x}{y}\Big)^{a + b}$

Hence, $\dfrac{\Big(x+\dfrac{1}{y}\Big)^a\Big(x-\dfrac{1}{y}\Big)^b}{\Big(y+\dfrac{1}{x}\Big)^a\Big(y-\dfrac{1}{x}\Big)^b} = \Big(\dfrac{x}{y}\Big)^{a + b}$.