If y = cos(3cos^-1x) then prove that d³y/dx³=24.

Dear student
$$\begin{gathered} \mathrm{y}=\cos \left(3{\cos }^{-1}\mathrm{x}\right) \\ \mathrm{Differentiating} \mathrm{wrt} \mathrm{x} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\left(-\sin \left(3{\cos }^{-1}\left(\mathrm{x}\right)\right)\right).\frac{\mathrm{d}}{\mathrm{dx}}\left[3{\cos }^{-1}\left(\mathrm{x}\right)\right] \\ =-3.\frac{\mathrm{d}}{\mathrm{dx}}\left[{\cos }^{-1}\mathrm{x}\right].\sin \left(3{\cos }^{-1}\mathrm{x}\right) \\ =-3\left(-\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\right)\sin \left(3{\cos }^{-1}\mathrm{x}\right) \\ =\frac{3\sin \left(3{\cos }^{-1}\mathrm{x}\right)}{\sqrt{1-{\mathrm{x}}^2}} \\ \mathrm{Differentiating} \mathrm{again} \mathrm{wrt} \mathrm{x} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{3\left({\displaystyle \frac{\mathrm{xsin}\left(3{\cos }^{-1}\mathrm{x}\right)}{\sqrt{1-{\mathrm{x}}^2}}}-3\cos \left(3{\cos }^{-1}\mathrm{x}\right)\right)}{1-{\mathrm{x}}^2} \\ \mathrm{Again} \mathrm{Differentiating} \mathrm{again} \mathrm{wrt} \mathrm{x} \\ \frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}=\frac{-24\sin \left(3{\cos }^{-1}\mathrm{x}\right)}{{\left(1-{\mathrm{x}}^2\right)}^{{\displaystyle \frac{3}{2}}}}+\frac{9{\mathrm{x}}^2\sin \left(3{\cos }^{-1}\mathrm{x}\right)}{{\left(1-{\mathrm{x}}^2\right)}^{\frac{5}{2}}}-\frac{27\mathrm{xcos}\left(3{\cos }^{-1}\mathrm{x}\right)}{{\left(1-{\mathrm{x}}^2\right)}^2} \end{gathered}$$
Please recheck your question , the answer is not 24.
Regards