The brackets are just there to show that the whole expression within is raised to the power $\displaystyle 4$. Are you aware of the general rule for differentiating something of the form $\displaystyle y=[f(x)]^n$?
Oh, you mean you want to solve it via a substitution. You don't have to solve it that way, which is why I was confused. Letting $\displaystyle t=\frac{6}{x}+4\ln(7x)$
And then using the fact that $\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}$ is the way to go. Why do you think you need to use the chain rule twice?
You start off correctly, but towards the end you get into a ball of confusion.
$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$
$\displaystyle y=8t^4$ and $\displaystyle t=6x^{-1}+4\ln(7x)$
$\displaystyle \frac{dy}{dt}=32t^3$
$\displaystyle \frac{dt}{dx}=-6x^{-2}+\frac{4}{x}$
Have another attempt.