Answer
$$y = {\tan ^{ - 1}}x{\text{ is a solution}}$$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}x \cr
& {\text{Calculate the first and second derivative}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x} \right] \cr
& y' = \frac{1}{{1 + {x^2}}} \cr
& y'' = - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{Substituting into }}y'' = - 2\sin y{\cos ^3}y \cr
& - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = - 2\sin \left( {{{\tan }^{ - 1}}x} \right){\left[ {\cos \left( {{{\tan }^{ - 1}}x} \right)} \right]^3} \cr
& - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = - 2\left( {\frac{x}{{\sqrt {{x^2} + 1} }}} \right){\left[ {\frac{1}{{\sqrt {{x^2} + 1} }}} \right]^3} \cr
& - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = - 2\left( {\frac{x}{{{{\left( {\sqrt {{x^2} + 1} } \right)}^4}}}} \right) \cr
& - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = - 2\left( {\frac{x}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right) \cr
& - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{The statement has been proved}} \cr} $$
