Correct process of finding double derivative Suppose I have obtained $\frac{d^2 y}{dt^2}=A$ and $\frac{dx}{dt}=B$.Can I square B and divide A by it to obtain $\frac{d^2 y}{dx^2}=\frac{A}{B^2}$ ?How is it wrong if it is , because I am getting different answers for $y=t^3$ and $x=t^2$ from this process and if I had obtained $\frac{dy}{dx}$ and then took its derivative to find the second derivative .
 A: 
Can I square B and divide A by it to obtain $\frac{d^2 y}{dx^2}=\frac{A}{B^2}$ ?

No, you can't.
You could be tempted to do this because the notation resembles (normal) powers/exponents, but they're not. More specifically:
$$\underbrace{\frac{\mbox{d}^2x}{\mbox{d}t^2}=\frac{\mbox{d}}{\mbox{d}t}\left(\frac{\mbox{d}x}{\mbox{d}t}\right)}_{\text{second derivative}} \ne \underbrace{\left(\frac{\mbox{d}x}{\mbox{d}t}\right)^2=\frac{\mbox{d}x}{\mbox{d}t}\frac{\mbox{d}x}{\mbox{d}t}}_{\text{square of the derivative}}$$


Suppose I have obtained $\frac{d^2 y}{dt^2}=A$ and $\frac{dx}{dt}=B$.Can I square B and divide A by it to obtain $\frac{d^2 y}{dx^2}=\frac{A}{B^2}$ ?

With $A$ and $B$ as above, you have:
$$\frac{A}{B^2}=\frac{\frac{d^2y}{dt^2}}{\left(\frac{dx}{dt}\right)^2}$$
and it 'ends' there. What you hope (?), is not valid:
$$\frac{\frac{d^2y}{dt^2}}{\left(\frac{dx}{dt}\right)^2} \color{red}{\ne} \frac{\frac{d^2y}{dt^2}}{\frac{dx^2}{dt^2}} \color{red}{"="} \frac{d^2y}{dx^2}$$
A: Your problem lies in the squaring of $B$.
Remember what that notation for a derivative actually means. It's essentially a short form of the "real" definition of the derivative, that is
$$\frac{df}{dx} = \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$$
Now you can see why the notation is that way, the $d$ symbols represent infinitesimal differences in $f$ and $x$, the numerator and denominator of the limit respectively. If you replace $B$ by the proper limit notation it will be obvious that squaring it (squaring the limit) is not the same as the double derivative, that is applying the derivative-limit method to the result of the derivative-limit method applied to $f$. 
Edit: Here's a counter-example for you. Let $f = -\cos x$.
$$\frac{d^2f}{dx^2} = \frac{d}{dx}\left[\frac{df}{dx}\right] = \frac{d}{dx}\left[\sin x\right] = \cos x \neq \sin^2 x = \left(\frac{df}{dx}\right)^2$$
A: The notation behind the second derivative is misleading (and, I would say, wrong).  The meaning behind the notation is that $d^2y$ is supposed to be a double-differentiation, and $dx^2$ is the quantity $dx$ (the differential of x) squared.  So, a more expansive way of writing it is this:
$$\frac{d(d(y))}{(d(x))^2}$$
The problem with this is that it is actually wrong.  To understand why, remember that the first derivative is a quotient:
$\frac{dy}{dx}$.  Therefore, taking the derivative of it will look like this:
$$\frac{d\left(\frac{dy}{dx}\right)}{dx}$$
Therefore, for the second derivative, since there is a quotient in the thing you are taking the differential of, you need to use the quotient rule.  Applying the quotient rule gives you the following:
$$\frac{\frac{dx\, d(dy) - dy\, d(dx)}{dx^2}}{dx}$$
Substituting $d^2x$ for $d(dx)$ and $d^2y$ for $d(dy)$ gives you the following for an algebraically manipulable second derivative:
$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
This is why your attempt does not work.  Using differentials algebraically is fine, but you have to use a notation that makes sense of differentials.  In your case, you would also need to know $\frac{d^2x}{dt^2}$ in order to make it all work (and the algebra is pretty ugly even then).
For more information about this, see the paper Extending the Algebraic Manipulability of Differentials.
A: The equation can be solved using the algebraic double derivative.
$$
A=\frac{d\left[\frac{dy}{dt}\right]}{dt}=\frac{d^2y}{dt^2}-\frac{dy\ d^2t}{dt^3}
$$
$$
B=\frac{dx}{dt}
$$
And thus, similarly
$$
B'=\frac{d^2x}{dt^2}-\frac{dx\ d^2t}{dt^3}
$$
If you desire to find $y''$, the double derivative of $y$ with respect to $x$, first write the algebraic double derivative
$$
y''=\frac{d^2y}{dx^2}-\frac{dy\ d^2x}{dx^3}
$$
One can manipulate the algebraic derivatives to get the following equation
$$
y''=\frac{A}{B^2}-B'B^3\frac{dy}{dx}
$$
Which recovers the expected $A/B^2$ and an additional term.
A: Before getting  the first equation you must have found $C=\frac {dy} {dt}$. Use $\frac {dy} {dx}=(\frac {dy} {dt})/(\frac {dx} {dt})$ to find $\frac {dy} {dx}$. Let $z=\frac {dy} {dx}$. Use $\frac {dz} {dx} =(\frac {dz} {dt})/(\frac {dx} {dt})$ to find the second derivative. 
