If $y={\cos }^{-1}\frac{9-x^2}{9+x^2}$ then $y^1(-1)$ is equal to

$y={\cos }^{-1}\frac{9-x^2}{9+x^2}={\cos }^{-1}\frac{1-(x/3{)}^2}{1+(x/3{)}^2}=$

$y=\left\{\begin{array}{ll}2{\tan }^{-1}(x/3)& 0\le x<\mathrm{\infty } \\ -2{\tan }^{-1}(x/3)& -\mathrm{\infty }<x\le 0\end{array}\right.$

In the neighbourhood of x=-1

$\frac{dy}{dx}=-\frac{2}{1+{\displaystyle \frac{x^2}{9}}}\left(\frac{1}{3}\right)=-\frac{6}{9+x^2}$

$y^1(-1)=\frac{-6}{10}=\frac{-3}{5}$