Basic question about higher derivative 
A curve in the $xy$
  -plane is given parametrically by
  $$x=t^2+2t \\ y=3t^4+4t^3$$
  for all $t>0$. What is the value of $\frac{d^2y}{dx^2}$ at $(8,80)$ ?

I have the following doubt:
We know that $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ Similarly, can we say that 
$$\frac{d^2y}{dx^2}=\frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}$$
 A: $$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} \\ \implies \frac{d^2 y}{dt^2} = \frac{d}{dt}\left(\frac{dy}{dx}\frac{dx}{dt}\right) = \frac{d^2y}{dx^2}\frac{dx}{dt}\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2} = \frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2+\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\frac{d^2x}{dt^2} \\ \implies \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}-\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^2}$$
A: The function $x(t) = t^2+2t$ has an inverse function when $t\gt 0$ (because $x(t)$ is continuous and monotonic when $t\gt0 $). So we can think of $t$ as a function of $x$.
We have: $${dy \over dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \tag{1}$$
To get the second derivative $\dfrac{d^2y}{dx^2}$, we differentiate (1) with respect to $x$ (keep in mind that $t$ is a function of $x$):
$$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left( \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right)= \dfrac{d}{\color{magenta}{dt}}\left( \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right)\dfrac{\color{magenta}{dt}}{dx}$$
But
$$\dfrac{d}{dt}\left( \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right) = \dfrac{\dfrac{dx}{dt}\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)-\dfrac{dy}{dt}\dfrac{d}{dt}\left( \dfrac{dx}{dt}\right)}{\left( \dfrac{dx}{dt}\right)^2}= \dfrac{\dfrac{dx}{dt}\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}\dfrac{d^2x}{dt^2}}{\left( \dfrac{dx}{dt}\right)^2}$$
$$\dfrac{dt}{dx} = \dfrac{1}{\dfrac{dx}{dt}}$$
So we find:
$$\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{dx}{dt}\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}\dfrac{d^2x}{dt^2}}{\left( \dfrac{dx}{dt}\right)^3}$$
