Answer
$$y'' = - \frac{{\sin y}}{{{{\left( {1 + \cos y} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& y + \sin y = x \cr
& {\text{differentiate both sides}} \cr
& \left( {y + \sin y} \right)' = \left( x \right)' \cr
& y' + y'\cos y = 1 \cr
& {\text{find }}y' \cr
& y'\left( {1 + \cos y} \right) = 1 \cr
& y' = \frac{1}{{1 + \cos y}} \cr
& {\text{find }}y'' \cr
& y'' = \left( {\frac{1}{{1 + \cos y}}} \right)' \cr
& {\text{quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& y'' = \frac{{\left( {1 + \cos y} \right)\left( 1 \right)' - \left( 1 \right)\left( {1 + \cos y} \right)'}}{{{{\left( {1 + \cos y} \right)}^2}}} \cr
& y'' = - \frac{{y'\sin y}}{{{{\left( {1 + \cos y} \right)}^2}}} \cr
& y'' = - \frac{{y'\sin y}}{{{{\left( {1 + \cos y} \right)}^2}}} \cr
& {\text{replace }}y' = \frac{1}{{1 + \cos y}} \cr
& y'' = - \frac{{\sin y}}{{{{\left( {1 + \cos y} \right)}^2}}}\left( {\frac{1}{{1 + \cos y}}} \right) \cr
& {\text{simplify}} \cr
& y'' = - \frac{{\sin y}}{{{{\left( {1 + \cos y} \right)}^3}}} \cr} $$
