Answer
$\frac{dy}{dx} = -\frac{2x(1+cscy)}{cscy}$
Work Step by Step
$x^{2} = \frac{coty}{1+cscy}$
Solution :
Differentiating implicitly :
$2x = \frac{(1+cscy)(-csc^{2}y)(\frac{dy}{dx})-(coty)(-cscycoty)(\frac{dy}{dx})}{(1+cscy)^{2}}$
$2x(1+cscy)^{2} = -cscy(cscy + csc^{2}y - cot^{2}y)\frac{dy}{dx}$
As $csc^{2}y -cot^{2}y = 1$
So
$2x(1+cscy)^{2} = -cscy(cscy + 1)\frac{dy}{dx}$
After simplification :
$2x(1+cscy) = -cscy\frac{dy}{dx}$
$\frac{dy}{dx} = -\frac{2x(1+cscy)}{cscy}$
