I was having a problem with this question, there are two methods I could have used, lets label these 1 and 2. Method 1 does not work but I have no idea why, however method 2 does work. The question is as follows:
For $x=\ln(2t+7)$ and $y=t^2,t>0$ find $\frac{d^2y}{dx^2}$ $$$$Method 1 does not work and I wanted to know why$$$$ Method 1: $$\frac{dx}{dt} = \frac{2}{2t+7}, \frac{dy}{dt}=2t$$ $$\frac{d^2x}{dt^2} = \frac{-4}{(2t+7)^2},\frac{d^2y}{dt^2}=2$$ So $$\frac{d^2y}{dx^2} =\frac{d^2y}{dt^2} \div \frac{d^2x}{dt^2}$$ $$= 2 \div \frac{-4}{(2t+7)^2} $$ $$ = \frac{-(2t+7)^2}{2}$$
Method 2: $$\frac{dx}{dt} = \frac{2}{2t+7}, \frac{dy}{dt}=2t$$ $$\frac{dy}{dx} =2t \times \frac{2t+7}{2} = 2t^2+7t$$ Implicit Differentiation $$\frac{d^2y}{dx^2} = (4t+7)\frac{dt}{dx}$$ $$\frac{d^2y}{dx^2} = (4t+7)\left(\frac{2t+7}{2}\right)$$ $$=\frac12 (4t+7)(2t+7)$$ This is the correct answer and so Method 1 clearly does not work but not sure why.
The mistake in your first method is that $d^2x\not =dx^2$. In the first case it indicates that you are derivating two time the function x, the second means that the derivation is in the variable x. It's very different.
Third method $$x=\ln(2t+7)\implies 2t+7=e^x\implies t=\frac{e^x-7}{2}\implies y=t^2=\frac{e^{2x}-14e^x+49}{4}$$
So $$\frac{dy}{dx}=\frac{2e^{2x}-14e^x}{4}$$ and $$\frac{d^2y}{dx^2}=\frac{4e^2x-14e^x}{4}=\frac{e^x(2e^x-7)}{2}=\frac{(2t+7)(4t+14-7)}{2}=\frac{(2t+7)(4t+7)}{2}$$