Tricky chain rule

Given that $x=t^2$ and $y = \ln v$ :

$\begin{aligned} \frac {dy}{dx} & = \frac {dy}{dv} \times \frac {dv}{dt} \times \frac {dt}{dx} = \frac 1v \times \frac {dv}{dt} \times \frac 1{2t} = \frac 1{2vt}\left(\frac {dv}{dt}\right) \\ \frac {d^2y}{dx^2} & = \frac d{dt} \left(\frac 1{2vt}\left(\frac {dv}{dt}\right)\right) \frac {dt}{dx} \\ & = \frac 1{4t} \left(\left(-\frac 1{v^2t}\left(\frac {dv}{dt}\right) - \frac 1{vt^2}\right)\frac {dv}{dt} +\frac 1{vt}\left(\frac {d^2v}{dt^2}\right)\right) \\ & = \frac 1{4xv}\left(\frac {d^2v}{dt^2}\right) - \frac 1{4xv^2}\left(\frac {dv}{dt}\right)^2 - \frac 1{4xvt}\left(\frac {dv}{dt}\right) \end{aligned}$

Therefore,

$\begin{aligned} 4x\frac{d^2y}{dx^2} + 4x\left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx} -1 & = 0 \\ \frac 1{v}\left(\frac {d^2v}{dt^2}\right) - \frac 1{v^2}\left(\frac {dv}{dt}\right)^2 - \frac 1{vt}\left(\frac {dv}{dt}\right) + 4x\left(\frac 1{2vt}\left(\frac {dv}{dt}\right)\right)^2 + \frac 2{2vt}\left(\frac {dv}{dt}\right) & = 1 \\ \frac 1{v}\left(\frac {d^2v}{dt^2}\right) & = 1 \\ \implies \frac {d^2v}{dt^2} & = \boxed v \end{aligned}$

$\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dx}\frac{dx}{dt}$

Note that:

$y=ln(v)$ $\Leftrightarrow$ $v=e^y$ $\Leftrightarrow$ $\frac{dv}{dy} = e^y = v$

$f' = \frac{dy}{dx}$ ; $\frac{dx}{dt}=2t$

Putting those together:

$\Rightarrow$ $\frac{dv}{dt} = vf'(2t)$

$\Rightarrow$ $\frac{d^2v}{dt^2} = \frac{d}{dt}(vf'(2t)) = \frac{dv}{dt}f'(2t) + v\frac{df'}{dt}(2t) + vf'\frac{d(2t)}{dt}$

$\Rightarrow$ $\frac{d^2v}{dt^2} = v(f'.2t)^2 +vf''(2t)^2 + 2vf' = 2v(2t^2f'^2 + 2t^2f'' +f')$

Now come back to the given equation:

$4x\frac{d^2y}{dx^2} + 4x(\frac{dy}{dx})^2 + 2\frac{dy}{dx} -1 =0$ $\Leftrightarrow$ $4t^2f'' +4t^2f'^2 +2f'=1$ $\Leftrightarrow$ $2t^2f'^2 + 2t^2f'' + f' =\frac{1}{2}$

Hence: $\frac{d^2v}{dt^2} = 2v(\frac{1}{2}) = v$