If $\sin x\cos y=\frac{1}{4},3\tan x=4\tan y$, then $\sin (x-y)=$

Given $\sin x\cos y=\frac{1}{4}$ .......1

$3\tan x=4\tan y$
$3\sin x\cos y=4\sin y\cos x$
from 1 

$3\times \frac{1}{4}=4\sin y\cos x$
$\sin y\cos x=\frac{3}{16}$ ............2
$\therefore \sin (x-y)=\sin x\cos y-\sin y\cos x$
$=\frac{1}{4}-\frac{3}{16}=\frac{1}{16}$