Как тут находят производную, я не понимаю...по каким формулам?

По формулам:
$1)\ \ [u+v]'=u'+v'\\\\ 2)\ \ (x^a)'=a*x^{a-1}\\\\ (x)'=(x^1)=1*x^{1-1}=1*x^0=1*1=1\\\\ (\frac{1}{x})'=(x^{-1})'=-1*x^{-1-1}=-x^{-2}=-\frac{1}{x^2}\\\\ 3)\ \ [c*f(x)]'=c*[f(x)]'=c*f'(x)\\\\ ------------------------------\\\\ (\frac{289}{x})'=(289*\frac{1}{x})'=289*(\frac{1}{x})'=289*(-\frac{1}{x^2})=-\frac{289}{x^2}\\\\ y(x)=-\frac{x^2+289}{x}=-(\frac{x^2}{x}+\frac{289}{x})=-(x+\frac{289}{x})=(-1)*(x+\frac{289}{x})\\\\ y'(x)=[(-1)*(x+\frac{289}{x})]'=(-1)*[x+\frac{289}{x}]'=\\\\$

$=-[(x)'+(\frac{289}{x})']=-[1+(-\frac{289}{x^2})]=\\\\ =-[1-\frac{289}{x^2}]=-[\frac{x^2}{x^2}-\frac{289}{x^2}]=-[\frac{x^2-289}{x^2}]=-\frac{x^2-289}{x^2}=\\\\ =\frac{-(x^2-289)}{x^2}=\frac{-x^2+289}{x^2}=\frac{289-x^2}{x^2}$