Answer
$${\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {2,2} \right)}} = - 2$$
Work Step by Step
$$\eqalign{
& {x^3} + {y^3} = 16 \cr
& \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{x^3}} \right) + \frac{d}{{dx}}\left( {{y^3}} \right) = \frac{d}{{dx}}\left( 1 \right) \cr
& {\text{find derivatives}} \cr
& 3{x^2} + 3{y^2}\frac{{dy}}{{dx}} = 0 \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& 3{y^2}\frac{{dy}}{{dx}} = - 3{x^2} \cr
& \frac{{dy}}{{dx}} = - \frac{{{x^2}}}{{{y^2}}} \cr
& \cr
& {\text{find the second derivative}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( { - \frac{{{x^2}}}{{{y^2}}}} \right) \cr
& {\text{use the quotient rule}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^2}\frac{d}{{dx}}\left[ {{x^2}} \right] - {x^2}\frac{d}{{dx}}\left[ {{y^2}} \right]}}{{{{\left( {{y^2}} \right)}^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^2}\left( {2x} \right) - {x^2}\left( {2y} \right)\frac{{dy}}{{dx}}}}{{{y^4}}} \cr
& \cr
& {\text{substitute }}\frac{{dy}}{{dx}} = - \frac{{{x^2}}}{{{y^2}}}{\text{ and simplify}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{2x{y^2} - 2{x^2}y\left( { - \frac{{{x^2}}}{{{y^2}}}} \right)}}{{{y^4}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{2x{y^2} + \frac{{2{x^4}}}{y}}}{{{y^4}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{2x{y^3} + 2{x^4}}}{{{y^5}}} \cr
& \cr
& {\text{Evaluate at the point }}\left( {2,2} \right) \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {2,2} \right)}} = - \frac{{2\left( 2 \right){{\left( 2 \right)}^3} - 2{{\left( 2 \right)}^4}}}{{{{\left( 2 \right)}^5}}} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {2,2} \right)}} = - 2 \cr} $$
