xtany = y-1

Here you must make use of product rule on the right side and implicit differentiation.

x * sec²(y) * y' + tan(y) = y'

Then just rearragnge to solve for y'

x * sec²(y) * y' - y' = -tan(y)

y'(xsec²(y) - 1) = -tan(y)

y' = -tan(y) / [xsec(y) - 1]

y = 1/sin(x-sinx)

Here, all that you have to keep in mind is that the derivative of sec(u) = sec(u)tan(u) * du/dx

The du/dx is not officially a part of the derivative of sec(x), it is just there to remind you of chain rule. (where u = x - sin(x))

y = sec(x - sin(x))

y' = sec(x-sin(x))tan(x - sin(x))(1 - cos(x))