$y=\large\frac{8^x}{x^8}$
$\;\;=8^xx^{-8}$
This can be differentiated by using the product rule .
$d(uv)=uv'+vu'$
Here $u=8^x,v=x^{-8}$
$u'=8^x\log 8,v'=-8x^{-9}$
$\large\frac{dy}{dx}$$=8^x.(-8x^{-9})+x^{-8}.\log 8$
$\quad=\large\frac{-8.8^x}{x^9}+\frac{8^x\log 8}{x^8}$
$\quad=\large\frac{8^{\large x}}{x^{\large 8}}$$[\log 8-\large\frac{8}{x}]$