Can we multiply between two Derivative operator? I don't know what should be the tile of my question sorry for that ..
I will explain here clearly ...
Suppose this is my integral :
$\int \frac{d^2x}{dt^2}dx$
I have to show that the integral is equal to $(\frac{dx}{dt})^2$
My steps :
$=\int \frac{d}{dt}(\frac{dx}{dt})dx$
$=\int d\frac{d^2x^2}{d^2t^2}$  Note:  Here I am multiplying  between the two operator like $dx \cdot dx=d^2x^2 $ and $dt \cdot dt=d^2t^2 $
$=(\frac{dx}{dt})^2$
So now my question is can we multiply like the way I did according to my note . If my procedure is wrong how can i show $(\frac{dx}{dt})^2$ from the integral

@WilliamBarnes has the correct answer, but I wanted to take some time to correct issues with your understandings of higher order differentials.  Additionally, I'll give a way to do this without the weird substitution.
The $d()$ is an operator, not a multiplier.  $dx$ is equivalent to saying $d(x)$, like you would say $\sin(x)$ or something.  $d^2(x)$ actually means applying the operator twice: $d(d(x))$.  Also, $dx^2$ actually means $(d(x))^2$.
However, the standard notation for the second derivative is actually misleading (or, I would say, wrong).  You shouldn't take the second derivative as meaning what it says.  Most people write it the second derivative of $x$ with respect to $t$ as you did: $\frac{d^2x}{dt^2}$.  However, this is fairly misleading as far as actual differentials go.  If you want manipulable differentials, you have to write the second derivative of $x$ with respect to $t$ as:
$$  \frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2} $$
Using this, you can now treat these as algebraically manipulable objects, keeping in mind the shorthands we are using (discussed in the first paragraph).  See my paper "Extending the Algebraic Manipulability of Differentials" for more information.
So, to do the integral, let's start (using the manipulable notation) by doing:
$$
I = \left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right)\,dx
$$
We can move one of the $dt$s to be with the $dx$, giving:
$$
I = \left(\frac{d^2x}{dt} - \frac{dx}{dt}\frac{d^2t}{dt}\right)\,\frac{dx}{dt}
$$
The left side (in parenthesis) is the differential of the first derivative ($\frac{dx}{dt}$).  Therefore, we will set it to be $dv$ for integration by parts. $\frac{dx}{dt}$ will be $u$ as well as $v$.
This means we can rewrite this as:
$$
I = \frac{dx}{dt}\frac{dx}{dt} - \int \frac{dx}{dt} \left(\frac{d^2x}{dt} - \frac{dx}{dt}\frac{d^2t}{dt}\right)
$$
You'll notice that the right-hand side of this is equivalent to our previous definition of $I$.  Therefore, we can rewrite this as:
$$
I = \frac{dx}{dt}\frac{dx}{dt} - I
$$
Adding $I$ to both sides yields:
$$
2I = \frac{dx}{dt}\frac{dx}{dt} \\
I = \frac{1}{2} \left(\frac{dx}{dt}\right)^2
$$
