Answer
$$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \sin 2y - 2y{{\sin }^2}y - y{{\cos }^2}}}{{{{\left( {1 + x\sin y} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& x\cos y = y \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {x\cos y} \right] = \frac{d}{{dx}}\left[ y \right] \cr
& - x\left( {\sin y} \right)y' + \cos y = y' \cr
& \cos y = y' + x\left( {\sin y} \right)y' \cr
& \frac{{\cos y}}{{1 + x\sin y}} = y' \cr
& \frac{{dy}}{{dx}} = \frac{{\cos y}}{{1 + x\sin y}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left( {1 + x\sin y} \right)y'\sin y - \cos y\left( {x\cos yy' + \sin y} \right)}}{{{{\left( {1 + x\sin y} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - y'\sin y - xy'{{\sin }^2}y - xy'{{\cos }^2}y - \cos y\sin y}}{{{{\left( {1 + x\sin y} \right)}^2}}} \cr
& {\text{Substituting }}y' = \frac{{\cos y}}{{1 + x\sin y}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \frac{{\sin y\cos y}}{{1 + x\sin y}} - \frac{{x{{\sin }^2}y\cos y}}{{1 + x\sin y}} - \frac{{x{{\cos }^3}y}}{{1 + x\sin y}} - \cos y\sin y}}{{{{\left( {1 + x\sin y} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \sin y\cos y - x{{\sin }^2}y\cos y - x{{\cos }^3}y - \cos y\sin y - x\cos y{{\sin }^2}y}}{{{{\left( {1 + x\sin y} \right)}^3}}} \cr
& {\text{Substituting }}y = x\cos y \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 2\sin y\cos y - 2y{{\sin }^2}y - y{{\cos }^2}}}{{{{\left( {1 + x\sin y} \right)}^3}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \sin 2y - 2y{{\sin }^2}y - y{{\cos }^2}}}{{{{\left( {1 + x\sin y} \right)}^3}}} \cr} $$
