differentiation

takin log both sides,

cotx logx + y(logtan^-1x) = 0
diffrentiating with respect to x,

-cosec²x.logx+1/x.cotx + dy/dx(1/tan^-1x.1/1+x²)=0

cosec²x.logx - cotx/x = dy/dx(1/tan^-1x.(1+x²))

dy/dx=tan^-1x(1+x²)(cosec²x.logx - cotx/x)

anchal , your first steo itself is wrong

u have wriiten cotx logx

it should be cotx logy

yup your method was very much correct.

no need for sorry , but never hurry like this in a exam....