Answer
$${\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {0,1} \right)}} = - \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& {y^3} + y = 2\cos x \cr
& \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{y^3}} \right) + \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {2\cos x} \right) \cr
& {\text{find derivatives}} \cr
& 3{y^2}\frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} = - 2\sin x \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& \left( {3{y^2} + 1} \right)\frac{{dy}}{{dx}} = - 2\sin x \cr
& \frac{{dy}}{{dx}} = - \frac{{2\sin x}}{{3{y^2} + 1}} \cr
& \cr
& {\text{find the second derivative}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{d}{{dx}}\left( {\frac{{2\sin x}}{{3{y^2} + 1}}} \right) \cr
& {\text{use the quotient rule}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\left( {3{y^2} + 1} \right)\frac{d}{{dx}}\left[ {2\sin x} \right] - 2\sin x\frac{d}{{dx}}\left[ {3{y^2} + 1} \right]}}{{{{\left( {3{y^2} + 1} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\left( {3{y^2} + 1} \right)\left( {2\cos x} \right) - 2\sin x\left( {6y} \right)\frac{{dy}}{{dx}}}}{{{{\left( {3{y^2} + 1} \right)}^2}}} \cr
& \cr
& {\text{substitute }}\frac{{dy}}{{dx}} = - \frac{{2\sin x}}{{3{y^2} + 1}}{\text{ and simplify}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\left( {3{y^2} + 1} \right)\left( {2\cos x} \right) - 2\sin x\left( {6y} \right)\left( { - \frac{{2\sin x}}{{3{y^2} + 1}}} \right)}}{{{{\left( {3{y^2} + 1} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\left( {3{y^2} + 1} \right)\left( {2\cos x} \right) + \frac{{24y{{\sin }^2}x}}{{3{y^2} + 1}}}}{{{{\left( {3{y^2} + 1} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{2\cos x{{\left( {3{y^2} + 1} \right)}^2} + 24y{{\sin }^2}x}}{{{{\left( {3{y^2} + 1} \right)}^3}}} \cr
& \cr
& {\text{Evaluate at the point }}\left( {0,1} \right) \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {0,1} \right)}} = - \frac{{2\cos \left( 0 \right){{\left( {3{{\left( 1 \right)}^2} + 1} \right)}^2} + 24\left( 1 \right){{\sin }^2}\left( 0 \right)}}{{{{\left( {3{{\left( 1 \right)}^2} + 1} \right)}^3}}} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {0,1} \right)}} = - \frac{{2{{\left( 4 \right)}^2} + 24\left( 0 \right)}}{{{{\left( {3{{\left( 1 \right)}^2} + 1} \right)}^3}}} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {0,1} \right)}} = - \frac{1}{2} \cr} $$
