How do you find points of inflection of (1-x^2)/x^3?

It's probably best to start by writing this function as $f \left(x\right) = {x}^{- 3} - {x}^{- 1}$. Then $f ' \left(x\right) = - 3 {x}^{- 4} + {x}^{- 2}$ and $f ' ' \left(x\right) = 12 {x}^{- 5} - 2 {x}^{- 3} = \setminus \frac{12 - 2 {x}^{2}}{{x}^{5}}$. Setting $f ' ' \left(x\right) = 0$ gives $x = \setminus \pm \setminus \sqrt{6}$ as possible inflection points. These are actual inflection points because $f ' ' \left(x\right)$ changes sign as $x$ increases through these values.