<s>[INST] $y=\frac{1}{x}$, show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find $\frac{d^2y}{dt^2}$ If $y=\frac{1}{x}$, where $x$ is a function of $t$ show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find an expression for $\frac{d^2y}{dt^2}$.
For $\frac{d^2y}{dt^2}$ I keep getting 
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\frac{dx}{dt}$$ 
but the correct answer is apparently 
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\left({\frac{dx}{dt}}\right)^2$$
 [/INST] You are misapplying the product rule. Note that
$$\begin{aligned}
\frac{d}{dt} \left( - \frac{1}{x^2} \frac{dx}{dt} \right) & = \left( \frac{d}{dt} \frac{(-1)}{x^2} \right) \frac{dx}{dt} - \frac{1}{x^2} \left( \frac{d}{dt} \frac{dx}{dt} \right) \\
 & = \left( \frac{2}{x^3} \cdot \frac{dx}{dt} \right) \frac{dx}{dt} - \frac{1}{x^2} \frac{d^2x }{dt^2} \\
 & = \frac{2}{x^3} \left( \frac{dx}{dt} \right)^2 - \frac{1}{x^2} \frac{d^2 x}{dt^2}.
\end{aligned}$$ 
You forgot to apply the chain rule to $- \frac{1}{x^2}$.
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