Answer
$$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\sin y}}{{{{\left( {\cos y - 1} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& x + y = \sin y \cr
& {\text{Use the implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {x + y} \right] = \frac{d}{{dx}}\left[ {\sin y} \right] \cr
& 1 + \frac{{dy}}{{dx}} = \cos y\frac{{dy}}{{dx}} \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& 1 = \cos y\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\cos y - 1}} \cr
& \frac{{dy}}{{dx}} = {\left( {\cos y - 1} \right)^{ - 1}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {{{\left( {\cos y - 1} \right)}^{ - 1}}} \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - {\left( {\cos y - 1} \right)^{ - 2}}\left( { - \sin y} \right)\frac{{dy}}{{dx}} \cr
& {\text{Where }}\frac{{dy}}{{dx}} = {\left( {\cos y - 1} \right)^{ - 1}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - {\left( {\cos y - 1} \right)^{ - 2}}\left( { - \sin y} \right){\left( {\cos y - 1} \right)^{ - 1}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - {\left( {\cos y - 1} \right)^{ - 3}}\left( { - \sin y} \right) \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\sin y}}{{{{\left( {\cos y - 1} \right)}^3}}} \cr} $$
