If $xy(y-x)=16$ then y has a minimum value when  $x=$

$y^{2}x-x^{2}y-16=0$

$\Rightarrow 2xy{y}^1+y^2-x^2{y}^1-2xy=0 \text{and }{y}^1=0$

$\Rightarrow y=2x$
$2x^3=16\Rightarrow x=2$

$\frac{da}{dx}=(\frac{-4}{{\sin }^{2}x}+\frac{1}{(1-\sin x{)}^2})\cos x=0$

$\Rightarrow \cos x\ne 0\Rightarrow \sin x=2/3$

$\Rightarrow \frac{d^{2}a}{d{x}^2}=45>0 \text{for }\sin x=2/3$

$\therefore a=\frac{4}{2/3}+\frac{1}{1-2/3}=6+3=9$