I'm reading Classical dynamics from Marion and Thornton, and in example 2.1 they solve the situation of a box moving down an inclined plane.
And the end of the exercise they have that:
$ \frac{dv}{dt} = g \sin(\theta) $
Now, they decide to find the velocity as a function of the displacement of the block and they do the following:
Multiply both sides by $2 \frac{dx}{dt} $:
$ 2 \frac{dx}{dt}\frac{dv}{dt} = 2 \frac{dx}{dt}g \sin(\theta) \tag{1}\label{1} $
$ \frac{d(\frac{dx}{dt}^2)}{dt} = 2 \frac{dx}{dt}g \sin(\theta) \tag{2}\label{2} $
And they proceed to remove $ dt $ and integrate both sides
It seems to me that they are doing a sort of chain rule from \eqref{1} to \eqref{2}:
$ \frac{d(\frac{dx}{dt}^2)}{dt} = 2\frac{dx}{dt}\frac{d(\frac{dx}{dt})}{dt} = 2\frac{dx}{dt}\frac{dv}{dt} $
But I don't understand the manipulation as both position and velocity can be derived with respect to time directly.
You are always free to introduce more variables if it helps you.
For example, if $y = t^6, \frac{dy}{dt} = 6t^5$. But you can also say $u = t^2, y = u^3,$ and write $$\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}$$ which becomes $\frac{dy}{dt} = 3u^2\cdot 2t = 3t^4 \cdot 2t = 6t^5$, the same answer. You have the option to do it in whichever way happens to be easier for you for a given problem.
In this one, they are essentially saying, "let's call $\frac{dx}{dt} = u$, so the left side is $u'$." Then they multiply by $2u$ on both sides. By choosing $u$ as a derivative they are ensuring that they can integrate the right side, and on the left they have created $\frac{du^2}{dt}$. It's a clever trick, I think.
Your chain rule is correct.