triggy got even better !

It is given that: $\dfrac {\sin{x}}{\sin{y}} = 3$ and $\dfrac {\cos{x}}{\cos{y}} = \dfrac {1}{2}$ .

$\Rightarrow \sin {y} = \frac {1}{3}\sin {x}\quad \cos {y} = 2 \cos {x}$

$\Rightarrow \sin^2 {y} = \frac {1}{9}\sin^2 {x}\quad \cos^2 {y} = 4 \cos^2 {x}$

$\Rightarrow \sin^2{y} + \cos^2{y} = \frac {1}{9}\sin^2 {x} + 4 \cos^2 {x} = 1$

$\Rightarrow \frac {1}{9}\sin^2 {x} + 4(1- \sin^2 {x}) = 1$

$\Rightarrow (4-\frac {1}{9})\sin^2 {x} = 4-1\quad \Rightarrow \frac {35}{9} \sin^2 {x} = 3 \quad \Rightarrow \sin^2 {x} = \frac {27}{35}$

$\Rightarrow \sin^2 {y} = \frac {1}{9}\sin^2 {x} = \frac {1}{9}\times \frac {27}{35} = \frac {3}{35}$

Now, we work on:

$\dfrac {\sin{2x}}{\sin{2y}} + \dfrac {\cos{2x}}{\cos{2y}} = \dfrac {2\sin{x}\cos{x}}{2\sin{y}\cos{y}} + \dfrac {1-2\sin^2{x}}{1-2\sin^2{y}} = 3\times \dfrac {1}{2} + \dfrac {1-\frac{54}{35}}{1-\frac{6}{35}}$

$\quad \quad \quad \quad \quad \quad \quad \space = \dfrac {3}{2} + \dfrac {35-54}{35-6} = \dfrac {3}{2} - \dfrac {19} {29} = \dfrac {87-38}{58} = \boxed{\frac {49}{58}}$