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?

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.
