Let $\xi=x-16 t$ and consider $$ u(x,t)=-12\frac{3+4\cosh(2\xi+24t)+\cosh(4\xi)}{\left(3\cosh(\xi-12t)+\cosh(3\xi+12t)\right)^2} $$
Now fix $\xi$; it is said that
It is said that $$ u(x,t)\sim -8\operatorname{sech}^2(2\xi-\frac{1}{2}\log 3)\textrm{ as }t\to\infty.~\tag{*} $$
(1) How can $\xi$ be fixed if $\xi=x-16 t$ and $t\to\infty$?
(2) How to see $(*)$?
As to (2), I already used $\cosh(x)=\frac{1}{2}(e^x+e^{-x})$ to write $$ u(x,t)=-12\frac{3+2\left(e^{2\xi+24t}+e^{-2\xi-24t}\right)+\frac{1}{2}\left(e^{4\xi}+e^{-4\xi}\right)}{\frac{1}{4}\left\{9\left(e^{2\xi-24t}+2+e^{-2\xi+24t}\right)+e^{6\xi+24t}+2+e^{-6\xi-24t}+6\left(e^{4\xi}\color{red}{+e^{-2\xi-24t}+e^{2\xi+24t}}+e^{-4\xi}\right)\right\}} $$
Does this help? I don't know how to continue.
To (1): Saying that $\xi$ is fixed while $\xi=x-16$ should be read as $x=x(t)=\xi+16t$ for that value of $\xi\in\mathbb{R}$, so $x$ is an implicit function.
To (2): If you factor out $e^{24t}$ both in the numerator and denominater and let them cancel out, the remaining terms do all vanish but $(-24)e^{2\xi}$ in the numerator and $\frac{9}{4}e^{-2\xi}+\frac{1}{4}e^{6\xi}$ in the denominator. That would be the leading term of the asymptotics. Now one has to show that this and $(*)$ coincide.