A calculus problem by Sabelo Sibeko

The second derivative of parametric equations is $\frac{d^2y}{dx^2} = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^3}$

For $\color{#20A900}\dfrac{dx}{dt}$ and $\color{#3D99F6}\dfrac{d^2x}{dt^2}$ , $\begin{array}{rl} {\color{#20A900}\dfrac{dx}{dt}} &= 1 - \dfrac{1}{t^2}\\ {\color{#3D99F6}\dfrac{d^2x}{dt^2}} &= \dfrac{2}{t^3} \end{array}$ For $\color{#D61F06}\dfrac{dy}{dt}$ and $\color{#69047E}\dfrac{d^2y}{dt^2}$ , $\begin{array}{rl} {\color{#D61F06}\dfrac{dy}{dt}} &= 1\\ {\color{#69047E}\dfrac{d^2y}{dt^2}} &= 0 \end{array}$ Thus, \begin{array}{rl} \dfrac{d^2y}{dx^2} &= \dfrac{{\color{#20A900}\dfrac{dx}{dt}}{\color{#69047E}\dfrac{d^2y}{dt^2}} - {\color{#3D99F6}\dfrac{d^2x}{dt^2}}{\color{#D61F06}\dfrac{dy}{dt}}}{\left({\color{#20A900}\dfrac{dx}{dt}}\right)^3}\\ &= \dfrac{\left(\color{#20A900}1 - \dfrac{1}{t^2}\right)\left({\color{#69047E}0}\right) - \left(\color{#3D99F6}\dfrac{2}{t^3}\right)\left(\color{#D61F06}1\right)}{\left({\color{#20A900}1 - \dfrac{1}{t^2}}\right)^3}\\ &= \dfrac{-\dfrac{2}{t^3}}{\left(1 - \dfrac{1}{t^2}\right)^3} \cdot \underbrace{\dfrac{\left(t^2\right)^3}{\left(t^2\right)^3}}_{\substack{\text{Multiply top} \\ \text{and bottom} \\ \text{by }\left(t^2\right)^3}}\\ &= \boxed{\dfrac{-2t^3}{\left(t^2 - 1\right)^3}} \end{array}