Answer
$\sin y = \left( {1 - \frac{1}{x}} \right)k + \frac{1}{2}$
Work Step by Step
$$\eqalign{
& {x^2}y' = k\sec y,{\text{ }}y\left( 1 \right) = \frac{\pi }{6} \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& {x^2}\frac{{dy}}{{dx}} = k\sec y \cr
& {\text{Separate the variables}} \cr
& \frac{1}{{\sec y}}dy = \frac{k}{{{x^2}}}dx \cr
& \cos ydy = k{x^{ - 2}}dx \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\cos y} dy = \int {k{x^{ - 2}}} dx \cr
& \sin y = \frac{{k{x^{ - 1}}}}{{ - 1}} + C \cr
& \sin y = - \frac{k}{x} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 1 \right) = \frac{\pi }{6} \cr
& \sin \left( {\frac{\pi }{6}} \right) = - \frac{k}{1} + C \cr
& \frac{1}{2} = - k + C \cr
& C = k + \frac{1}{2} \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& \sin y = - \frac{k}{x} + k + \frac{1}{2} \cr
& \sin y = \left( {1 - \frac{1}{x}} \right)k + \frac{1}{2} \cr} $$
