Answer
$$\frac{dy}{dx}=\frac{-2x\csc x^2\cot x^2(1-\cot x^2)-(1+\csc x^2)2x\csc ^2x^2}{(1-\cot x^2)^2}$$
Work Step by Step
$$\frac{dy}{dx}=\Big(\frac{1+\csc x^2}{1-\cot x^2}\Big)'=
\frac{(1+\csc x^2)'(1-\cot x^2)-(1+\csc x^2)(1-\cot x^2)'}{(1-\cot x^2)^2}=
\frac{(-\csc x^2\cot x^2\cdot(x^2)')(1-\cot x^2)-(1+\csc x^2)(\csc^2x^2\cdot(x^2)')}{(1-\cot x^2)^2}=
\frac{-2x\csc x^2\cot x^2(1-\cot x^2)-(1+\csc x^2)2x\csc ^2x^2}{(1-\cot x^2)^2}$$
