If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$ 
Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and  $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$

My attempts:
Using the given condition and double angle formula, 
$\dfrac{\sin2x}{\sin2y} = \dfrac{3}{2}$
Now I am struggling to find the value of $\dfrac{\cos2x}{\cos2y}$
The simplest form I could reach was: 
$\dfrac{2\cos^2x-1}{8\cos^2x-1}$
How do I continue from here? Simpler methods to solve the problem are welcome.
PS: The answer is $\dfrac{49}{58}$  

hint
$$\sin (y)=\frac {1}{3}\sin(x) $$
$$\cos (y )=2\cos (x) $$
$$\implies \sin^2(x )+36\cos^2 (x)=9$$
$$\implies  35\cos^2 (x)=8$$
$$\implies \cos^2 (x)=\frac {8}{35} $$
the second expression is
$$\frac {16-35}{64-35}=-19/29$$
the final result is
$$3/2-19/29=49/58$$
