Answer
$f''\left( \theta \right) = \frac{{2\sin \theta - {\theta ^2}\sin \theta - 2\theta \cos \theta }}{{{\theta ^3}}}$
Work Step by Step
$$\eqalign{
& f\left( \theta \right) = \frac{{\sin \theta }}{\theta } \cr
& {\text{Differentiate }}f\left( \theta \right){\text{ to find }}f'\left( \theta \right) \cr
& f'\left( \theta \right) = \frac{d}{{d\theta }}\left[ {\frac{{\sin \theta }}{\theta }} \right] \cr
& {\text{Use the quotient rule}} \cr
& f'\left( \theta \right) = \frac{{\theta \frac{d}{{d\theta }}\left[ {\sin \theta } \right] - \sin \theta \frac{d}{{d\theta }}\left[ \theta \right]}}{{{\theta ^2}}} \cr
& f'\left( \theta \right) = \frac{{\theta \cos \theta - \sin \theta }}{{{\theta ^2}}} \cr
& \cr
& {\text{Differentiate }}f'\left( \theta \right){\text{ to find }}f''\left( \theta \right) \cr
& f''\left( \theta \right) = \frac{d}{{d\theta }}\left[ {\frac{{\theta \cos \theta - \sin \theta }}{{{\theta ^2}}}} \right] \cr
& {\text{Use the quotient rule}} \cr
& f''\left( \theta \right) = \frac{{{\theta ^2}\frac{d}{{d\theta }}\left[ {\theta \cos \theta - \sin \theta } \right] - \left( {\theta \cos \theta - \sin \theta } \right)\frac{d}{{d\theta }}\left[ {{\theta ^2}} \right]}}{{{{\left( {{\theta ^2}} \right)}^2}}} \cr
& {\text{Use the product rule for }}\frac{d}{{d\theta }}\left[ {\theta \cos \theta } \right] \cr
& f''\left( \theta \right) = \frac{{{\theta ^2}\left[ {\theta \left( { - \sin \theta } \right) + \cos \theta - \cos \theta } \right] - \left( {\theta \cos \theta - \sin \theta } \right)\left( {2\theta } \right)}}{{{\theta ^4}}} \cr
& {\text{Simplify}} \cr
& f''\left( \theta \right) = \frac{{{\theta ^2}\left( { - \sin \theta } \right) - 2\theta \cos \theta + 2\sin \theta }}{{{\theta ^3}}} \cr
& f''\left( \theta \right) = \frac{{2\sin \theta - {\theta ^2}\sin \theta - 2\theta \cos \theta }}{{{\theta ^3}}} \cr
& \cr
& {\text{Summary}} \cr
& f'\left( \theta \right) = \frac{{\theta \cos \theta - \sin \theta }}{{{\theta ^2}}}{\text{ }} \cr
& {\text{and }} \cr
& f''\left( \theta \right) = \frac{{2\sin \theta - {\theta ^2}\sin \theta - 2\theta \cos \theta }}{{{\theta ^3}}} \cr} $$
