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