I think I'm on the right track here, but am not sure if I can tidy it up any better and also if its the correct answer

$\displaystyle

\begin{array}{l}

y' = y^2 \sin x \\

\frac{{dy}}{{dx}} = y^2 \sin x \\

\int {\frac{{dy}}{{y^2 }} = \int {(\sin x)dx} } \\

\int {(y^{ - 2} )dy = } \int {(\sin x)dx} \\

\frac{{y^{ - 1} }}{{ - 1}} = - \cos x + c \\

\frac{{ - 1}}{y} = - \cos x + c \\

y = \frac{1}{{\cos x + c}} \\

\end{array}

$

Where c is my constant.