<s>[INST] Finding $\frac{d^2y}{dx^2}$ for a curve given by $(x,y)=(t^2+2t,3t^4+4t^3)$. Why does my method yield the incorrect answer? The other day I saw what was, seemingly, a fairly simple question.

A curve in the $xy$-plane is given parametrically by the equations:
$$\begin{align}
x &=\phantom{3}t^2+2t \\
y &=3t^4+4t^3
\end{align}$$
for all $t>0$. Find the value of $\frac{d^2y}{dx^2}$ at $(8,80)$.

My method to obtain the correct answer is as follows:
$$\frac{dx}{dt}=2t+2 \tag1$$
$$(\frac{dx}{dt})^2=\frac{dx^2}{dt^2}=4t^2+8t+4 \tag2$$
$$\frac{dy}{dt}=12t^3+12t^2\implies\frac{d^2y}{dt^2}=36t^2+24t \tag3$$
$$\frac{\frac{d^2y}{dt^2}}{\frac{dx^2}{dt^2}}=\frac{d^2y}{dt^2}\frac{dt^2}{dx^2}=\frac{d^2y}{dx^2}=\frac{36t^2+24t}{4t^2+8t+4}\tag4$$
$(8,80)$ occurs when $t=2$. Plugging this value of $t$ into the equation above yields an answer of $\frac{16}{3}$. The answer, however, should be $4$.
Where did I go wrong and why, exactly, did this particular method fail to give the correct answer?
 [/INST] As mentioned in the comments, you cannot divide the individual second derivatives to get the answer. The correct way of doing this is as follows:
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}$$
we have
$$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{12t^2(t+1)}{2(t+1)} = 6t^2$$
and
$$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2(t+1)}$$
which means that the final answer would be
$$\frac{d^2y}{dx^2} = \frac{3t^2}{t+1}$$
substituting $t=2$ yields $4$, which is the correct answer.
</s>