Two methods of finding out $\frac{d^2y}{dx^2}$ when $\frac{dy}{dt} = 4t-2$ and $\frac{dx}{dt} = 2t + 3$ Find $\frac{d^2y}{dx^2}$ given that $\frac{dy}{dt} = 4t-2$ and $\frac{dx}{dt} = 2t + 3$ at $t=2$
I tried two methods for this question, both of which give me different answers. Which one is wrong and why so?
Method $1$: $\frac{dy}{dt} = 4t-2$ hence $\frac{d^2y}{dt^2} = 4$, similarly $\frac{d^2x}{dt^2} = 2$ therefore, $\frac{d^2y}{dx^2} = 2$
Method $2$: $\frac{dy}{dx} = \frac{4t-2}{2t+3}$ therefore $\frac{d^2y}{dx^2} = \frac{(2t+3)4 - (4t-2)2}{(2t+3)^2}.\frac{dt}{dx} = \frac{16}{(2t+3)^2}.\frac{1}{2t+3} = \frac{16}{7^3}$
I can't understand why one method is wrong while the other is right, however they have very different answers. What am I doing wrong?

Method #2 is the correct one, using the formula...
$$\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} $$
which is valid.
Method #1 is using a formula ...
$$\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} \frac{d^2t}{dx^2} \tag{???}$$
which is not valid. To see this consider the case where $t=x$,for which
$\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2}$, but $(???)$ would give $\frac{d^2y}{dx^2} = 0$
