#### Answer

The Derivative is:
$y'=\csc x - 2\csc^3 x$

#### Work Step by Step

$y=\frac{\cos x}{sin^{2}x}$
Applying trigonometric identities
$y=\frac{\cot x}{\sin x}$
$y=\cot x\csc x$
Applying Derivative Rules:
$y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$
$y'=(-\csc^2 x)(\csc x)+(-\csc x\cot x)(\cot x)$
$y'=-\csc^3 x-\csc x\cot^2 x$
Applying trigonometric identities
$\cot^2 x = -1+\csc^2 x$
$y'=-\csc^3 x-\csc x(-1+\csc^2 x)$
$y'=-\csc^3 x + \csc x - \csc^3 x$
$y'=\csc x - 2\csc^3 x$