The quotient #\frac[f(x)}{g(x)}# of two derivable functions #f(x)# and #g(x)# is derivable for all #x# in which #g(x)\ne 0#. And its derivative is equal to:

#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{(g(x))^2}#

So, going back to the exercise we have:

#f(x)=\sin(x)#

#g(x)=x^8#

Thus:

#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{x^8\cdot (-\sin(x)) - \cos(x)\cdot 8x^7}{(x^8)^2}#

#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^8\sin(x)-8x^7\cos(x)}{x^16}#

Factoring:

#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^7(x\sin(x)+8\cos(x))}{x^16}#

Reducing #frac{x^7}{x^16}# we finally get:

#frac{d}{dx}(\frac[f(x)}{g(x)})=-\frac{x\sin(x)+8\cos(x)}{x^9}#

For the second derivative your functions #f(x)# and #g(x)# will be:

#f(x)=-(x\sin(x)+8\cos(x))#

#g(x)=x^9#

Try doing it!