Use implicit differentiation to show
dy/dx = [(1xy)cot(y)]/[x^2 cot(y) + xcsc^2(y)]
if xy = ln(x cot y).
1 answer

from xy = ln(x coty)
x dy/dx + y = (x(csc^y dy/dx) + coty)/(x coty)
crossmultiply
(x^2)(coty)dy/dx + xycoty = x(csc^2y)dy/dx + coty
(x^2)(coty)dy/dx + x(csc^2y)dy/dx = coty  xycoty
dy/dx((x^2)(coty) + x(csc^2y)) = coty(1  xy)
dy/dx = (1xy)coty/[(x^2)(coty)dy/dx + x(csc^2y)]