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 $\frac{d^2 t}{dx^2}$?


The solution given by Deepak follows the usual way that the second derivative (parametric) function would be calculated for this curve.  The way you followed creates somewhat more work, but will finally achieve the same result.
This approach is somewhat similar to that of Clarinetist, but takes advantage of a particular feature of the parametric curve functions; it also proceeds entirely by implicit differentiation.  We may write $$ \ x \ = \ t^2 + 2t \  = \ (t+1)^2 - 1 \ \ , \ \ y \ = \ 3t^4 + 4t^3 \ = \ t^3 · (3t+4) \ = \ t^3 · [ (3t+3) + 1] \ \ . $$  We will then substitute $ \ u = t+1 \ \ ; $  consequently,
$$ x + 1 \ = \ u^2 \ \ \Rightarrow \ \ \frac{d}{dx} \ [x + 1] \ = \ \frac{d}{dx} \ [u^2] \ \ \Rightarrow \ \ 1 \ = \ 2u \ \frac{du}{dx} $$
$$ \Rightarrow \ \ \frac{d}{dx} \ [1] \ = \ \frac{d}{dx} \left[2u \ \frac{du}{dx} \right] \ \ \Rightarrow \ \ 0 \ = \ 2·\left(\frac{du}{dx} \right)^2 \ + \ 2u· \frac{d^2u}{dx^2} \ \ \Rightarrow \ \ \frac{d^2u}{dx^2} \ = \ -\frac{1}{u}·\left(\frac{du}{dx} \right)^2 \ \ . $$
For the other curve equation, we have
$$ y \ = \ t^3 · [ 3(t+1) + 1] \ \rightarrow \ (u-1)^3 · ( 3u + 1)  $$
$$ \Rightarrow \ \ \frac{dy}{du} \ = \ 3·(u-1)^2 · ( 3u + 1) \ + \ (u-1)^3 · 3 \ = \ 3·(u-1)^2 · [(3u+1) + (u-1)] $$ $$ = \ 12u·(u-1)^2 \ \ . $$
It follows that $$ \ \ \frac{dy}{dx} \ = \ \frac{dy}{du} · \frac{du}{dx} \ = \ 12u·(u-1)^2 \ · \ \frac{1}{2u} \ = \ 6·(u-1)^2 \ \rightarrow \ 6t^2 \ \ . $$
[We will remark upon this later.]
For the second derivative functions, we obtain
$$  \frac{d^2y}{du^2} \ = \  \frac{d}{du} \ [ 12u·(u-1)^2 ] \ = \ 12 ·(u-1)^2 \ + \ 12u · 2 \ (u-1) $$ $$ = \ 12 ·(u-1) · [(u-1) + 2u] \ = \ 12 ·(u-1) ·  (3u-1) \ \ ; $$
$$  \frac{d^2y}{dx^2} \ = \  \frac{d}{dx} \ \left[\frac{dy}{dx} \right] \ = \ \frac{d}{dx} \ \left[\frac{dy}{du} · \frac{du}{dx} \right] \ = \ \left(\frac{d}{du} \  \left[\frac{dy}{du} \right]  · \frac{du}{dx} \right) · \frac{du}{dx} \ + \ \frac{dy}{du} \ · \frac{d}{dx} \  \left[ \frac{du}{dx} \right] $$
$$  = \  \frac{d^2y}{du^2}  · \left(\frac{du}{dx} \right)^2  \ + \ \frac{dy}{du} \ ·  \frac{d^2u}{dx^2}   $$
[similar to your result, since $ \ \frac{du}{dx} = \frac{dt}{dx} ] $
(bringing in the expressions found above)
$$ = \  \frac{d^2y}{du^2}  · \left(\frac{du}{dx} \right)^2  \ + \ \frac{dy}{du} \ ·  \left[-\frac{1}{u}·\left(\frac{du}{dx} \right)^2 \right] \ = \ \left[\frac{d^2y}{du^2} \ - \ \frac{1}{u} · \frac{dy}{du} \right]  ·   \left(\frac{du}{dx} \right)^2    $$
$$  = \ \left[12·(u-1)·(3u-1) \ - \ \frac{1}{u} · 12u·(u-1)^2 \right]  ·   \left(\frac{1}{2u} \right)^2   $$
$$  = \ 12·(u-1) \ · \ [ (3u-1) \ - \  (u-1) ]  ·   \left(\frac{1}{2u} \right)^2 \ =  \ 12·(u-1) · 2u · \left(\frac{1}{2u} \right)^2 $$
$$ = \ \frac{6·(u-1)}{u} \ \ \rightarrow \ \ \frac{6t}{t \ +  \ 1} \ \ . $$
So this may not be the quickest route to the desired second derivative, but it is effective.
In the graphs at the start of this post, we see that this is a very narrow "hairpin" curve, with two "arms" almost entirely in the first quadrant, with a "turn" close to the origin [shown at lower right].  A curious feature about the curve (which is what interested me in this problem) is that the first derivative parametric function, $ \ 6t^2 \ , $ betrays no hint that there is something peculiar about the turn; the function is defined everywhere and is continuous, and the slope of the curve is always positive, except where it is zero at the origin.  However, the second derivative function is not defined at $ \ t = -1 \   $ (the point $ \ (-1,-1) \ ) , $ changes discontinuously from positive for $ \ t < -1 \ $ to negative for $ \ -1 < t < 0 \ , $ then reaches an inflection point at the origin $ \ (t=0) \ $ and returns to "upward" concavity for $ \ t > 0 \ \ . $
