Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 154: 56

$$y'' = -y^{-3}$$

Given$$x^{2}+y^{2}=1 $$ We differentiate with respect to $x$ \begin{align*} 2x+2yy'&=0\\ y'&=\frac{-x}{y} \end{align*} and \begin{align*} y''&= \frac{-y+xy'}{y^2}\\ &= \frac{-y+x\frac{-x}{y}}{y^2}\\ &= -\frac{x^2+y^2}{y^3} \\ & = -\frac{-1}{y^3} \\ & = -y^{-3} \end{align*}