A question about change of variables in an ODE or in general I have a question about differentiation in this question here:
differential equation Cauchy-Euler

I understand that it uses product rule to go from the 2nd line to the 3rd line (where the arrow point from and pointing to in the picture). But I was not sure how to get the second term in the 3rd line (the one being circled). I think I understand how to get the first term, but not able to figure out how to get the second term.
I guess the second term is $\frac{1}{x} * \frac{d}{dx}(\frac{dy}{dt})$, so how is it being manipulated to become that term inside the circle there? Are we allowed to just add say: $dp$ in nominator, and $dp$ in denominator however we wish?
Say if I have a term like this: $\frac{dx}{dz}$, can I just add $dp$ to that fraction anyway I wish as long as bottom and top "cancel outs"? like: $\frac{dx}{dz}=\frac{dx}{dp}\frac{dp}{dz} $?
Is it what is being done in that part inside the circle? (is it actually valid to do this if this is actually being done?)
 A: I guess this is one of the cases in which Leibniz notation doesn't really help. In the link you provided, $t = \log x$, so $t = t(x)$ is a function of $x$. So it's better to write $\frac{dy}{dt} = y^\prime (t(x))$. Then, by the chain rule, $$\frac{d}{dx} \left( \frac{dy}{dt} \right) = \frac{d}{dx} y^\prime (t(x)) = y^{\prime \prime} (t(x)) t^\prime (x) = \frac{d^2 y}{dt^2} \frac{dt}{dx}$$
A: Do the same for $y''$:
$$
\begin{align}
y''=&\frac {d^2y}{dx^2}=\frac {d}{dx}\left (\frac {dy}{dx} \right )\\
y''=&\frac {d}{dx}\left ( \frac {dy}{dt} \frac 1 x  \right ) \\
\end{align}
$$
Here you have to differentiate a product of two functions so you apply the rule :
$$(fg)'=f'g+fg'$$
$$y''=\dfrac {dy}{dt}\frac {d}{dx}\left (  \frac 1 x  \right ) 
+\frac 1 x  
\frac {d}{dx}\left ( \frac {dy}{dt}   \right ) 
$$
The first term is easy to calculate for the second term you have to apply the chain rule;
$$\dfrac {d}{dx}=\dfrac {d}{dt}\dfrac {dt}{dx}=\dfrac 1x\dfrac {d}{dt}$$
Since $t=\ln x \implies \dfrac {dt}{dx}=\dfrac 1x$.
A: I think my following work will be helpful to clear your doubt and make the work understandable.
\begin{align} y''=&\frac {d^2y}{dx^2}=\frac {d}{dx}\left (\frac {dy}{dx} \right )\\
=&\frac {d}{dx}\left ( \frac {dy}{dt} \frac 1 x  \right ) \\
=&\frac 1 x \cdot \frac {d}{dx}\left ( \frac {dy}{dt} \right )+\frac {dy}{dt}\cdot\frac {d}{dx}\left (\frac 1 x  \right )\\
=&\frac 1 x \cdot \frac {d}{dt}\left ( \frac {dy}{dt} \right )\cdot\frac {dt}{dx}-\frac 1 {x^2}\cdot\frac {dy}{dt}\\
=&\frac 1 {x^2} \cdot \frac {d}{dt}\left ( \frac {dy}{dt} \right )-\frac 1 {x^2}\cdot\frac {dy}{dt}\\
=&\frac 1 {x^2} \cdot \frac {d^2y}{dt^2} -\frac 1 {x^2}\cdot\frac {dy}{dt}\\
x^2y''=&\frac {d^2y}{dt^2}-\frac {dy}{dt}\\
\end{align}
Note: As $~x=e^t,~~ t=\ln x~$ and therefore $~{dt}/{dx}=1/x~.$
Edit:

Op's query: For the first term in the 4th line, is it possible to know how you get $\frac{d}{dt}(\frac{dy}{dt})\frac{dt}{dx}$ ? Could you elaborate little more ?

Explanation: Clearly here $~y=y(x)~$ i.e., $~y~$ is a function of $~x~$ and as $~x=e^t$ $(dx/dt=e^t\ne 0)$, so $~x~$ is again a function of $~t~$. Therefore $$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{dy}{dt}\cdot\dfrac{dt}{dx}\tag{*}$$
Now replace $~y~$ by $~\frac{dy}{dt}~,$ we have
$$\dfrac{d}{dx}\left(\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)\cdot\dfrac{dt}{dx}$$

*

*For the proof of the equality $(*)$, see Does the reciprocal of  $dx/dy$  equal  $dy/dx$ ?
