Answer
$$ y'= -\frac{\sin^{-1}x +\cos^{-1}x }{(\sin^{-1}x)^2
\sqrt{1-x^2}}.$$
Work Step by Step
Since $ y=\frac{\cos^{-1}x}{\sin^{-1}x}$, then by the qoutient rule, the derivative $ y'$ is given by
$$ y'= \frac{\sin^{-1}x\left(\frac{-1}{\sqrt{1-x^2}}\right)-\cos^{-1}x\left(\frac{1}{\sqrt{1-x^2}}\right)}{(\sin^{-1}x)^2}\\=-\frac{\sin^{-1}x +\cos^{-1}x }{(\sin^{-1}x)^2
\sqrt{1-x^2}}.$$
