Solve:

$\cos \left(\sin ^{-1} x\right)=\frac{1}{6}$

$\Rightarrow \cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right)=\frac{1}{6}$

$\Rightarrow \sqrt{1-x^{2}}=\frac{1}{6}$

$\Rightarrow 1-x^{2}=\frac{1}{36}$

$\Rightarrow 1-\frac{1}{36}=x^{2}$

$\Rightarrow x^{2}=\frac{35}{36}$

$\Rightarrow x=\pm \frac{\sqrt{35}}{6}$