Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?

By implicit differentiation on $x = t^2 + 2t$, one obtains
$$1 = 2t \cdot \dfrac{\text{d}t}{\text{d}x} + 2\dfrac{\text{d}t}{\text{d}x} = \dfrac{\text{d}t}{\text{d}x}\left(2t+2 \right)$$
hence
$$\dfrac{\text{d}t}{\text{d}x} = \dfrac{1}{2t+2}\text{.}$$
Differentiating a second time on the equation $x = t^2 + 2t$, one obtains by an application of the product rule
$$0 = 2\left(t \cdot \dfrac{\text{d}^2t}{\text{d}x^2} + 1 \cdot \dfrac{\text{d}t}{\text{d}x}  \right) + 2\text{.}$$
It follows from the above that
$$\dfrac{\text{d}^2t}{\text{d}x^2} = \dfrac{\dfrac{-2}{2} - \dfrac{\text{d}t}{\text{d}x}}{t} = \dfrac{-1 - \dfrac{\text{d}t}{\text{d}x}}{t} = -\left(\dfrac{1 + \dfrac{\text{d}t}{\text{d}x}}{t} \right)\text{.}$$
Substitute $\dfrac{\text{d}t}{\text{d}x}$ as found earlier, and you're done.
Note that you will have to solve for $t$ to compute this quantity. (Hint: make use of the fact that $x = t^2 = 2t$ and that you have the point $(x, y) = (8, 80)$.)
