Chapter 2 - The Derivative - 2.7 Implicit Differentiation - Exercises Set 2.7 - Page 166: 16

$y'' = \frac{{4}}{{{{\left( {x + 2y} \right)}^3}}}$

$$\eqalign{ & xy + {y^2} = 2 \cr & {\text{differentiate both sides}} \cr & \left( {xy} \right)' + \left( {{y^2}} \right)' = \left( 2 \right)' \cr & {\text{product rule}} \cr & xy' + y + 2yy' = 0 \cr & xy' + 2yy' = - y \cr & {\text{find }}y' \cr & y' = \frac{{ - y}}{{x + 2y}} \cr & {\text{find }}y'' \cr & y'' = - \left( {\frac{y}{{x + 2y}}} \right)' \cr & {\text{quotient rule }} \cr & \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr & y'' = - \frac{{\left( {x + 2y} \right)\left( y \right)' - y\left( {x + 2y} \right)'}}{{{{\left( {x + 2y} \right)}^2}}} \cr & y'' = - \frac{{\left( {x + 2y} \right)y' - y\left( {1 + 2y'} \right)}}{{{{\left( {x + 2y} \right)}^2}}} \cr & y'' = - \frac{{xy' + 2yy' - y - 2yy'}}{{{{\left( {x + 2y} \right)}^2}}} \cr & {\text{simplify}} \cr & y'' = - \frac{{xy' - y}}{{{{\left( {x + 2y} \right)}^2}}} \cr & {\text{replace }}y' = \frac{{ - y}}{{x + 2y}} \cr & y'' = - \frac{{x\left( {\frac{{ - y}}{{x + 2y}}} \right) - y}}{{{{\left( {x + 2y} \right)}^2}}} \cr & y'' = - \frac{{ - xy - xy - 2{y^2}}}{{{{\left( {x + 2y} \right)}^3}}} \cr & y'' = - \frac{{ - 2xy - 2{y^2}}}{{{{\left( {x + 2y} \right)}^3}}} \cr & y'' = \frac{{2xy + 2{y^2}}}{{{{\left( {x + 2y} \right)}^3}}} \cr & {\text{replace }}xy+y^2=2 \cr & y'' = \frac{{2(xy + {y^2})}}{{{{\left( {x + 2y} \right)}^3}}}=\frac{4}{(2x+y)^3} \cr} $$