Let y = f(x) = ^ 3 ((π)/(3)( ( π 3 2 (- 4 x ^ 3 + 5 x ^ 2 +...

The given equation can be written as

$y = \sin^{3}$\left(\dfrac{\pi}{3}cosg\left(x\right)\right)

where,

$g\left(x\right) = \dfrac{\pi}{3\sqrt{2}}{\left(- 4{x}^{3} + 5{x}^{2} + 1\right)}^{3 / 2}$

$⇒ {g}^{'}\left(x\right) = \dfrac{\pi}{2\sqrt{2}}{\left(- 4{x}^{3} + 5{x}^{2} + 1\right)}^{1 / 2}\left(- 12{x}^{2} + 10x\right)$

$⇒ {g}^{'}\left(1\right) = \dfrac{\pi\sqrt{2}}{2\sqrt{2}}\left(- 2\right) = - \pi$

And,

$g\left(1\right) = \dfrac{2\pi}{3} = \left(\pi - \dfrac{\pi}{3}\right)$

Now,

${y}^{'} = 3\sin^{2}\left(\dfrac{\pi}{3}cosg\left(x\right)\right) \times \cos \left(\dfrac{\pi}{3}cosg\left(x\right)\right) \times \dfrac{\pi}{3}\left(- sing\left(x\right)\right){g}^{'}\left(x\right)$

${y}^{'}\left(1\right) = 3\sin^{2}\left(- \dfrac{\pi}{6}\right) \cdot \cos \left(\dfrac{\pi}{6}\right) \cdot \dfrac{\pi}{3}\left(- sin\dfrac{2\pi}{3}\right){g}^{'}\left(1\right)$

${y}^{'}\left(1\right) = \dfrac{3}{4} \cdot \dfrac{\sqrt{3}}{2} \cdot \dfrac{\pi}{3}\left(\dfrac{- \sqrt{3}}{2}\right)\left(- \pi\right) = \dfrac{3{\pi}^{2}}{16}$

$y\left(1\right) = \sin^{3}\left(\dfrac{\pi}{3}cos\dfrac{2\pi}{3}\right) = - \dfrac{1}{8}$

Therefore,

$2{y}^{'}\left(1\right) + 3{\pi}^{2}y\left(1\right) = 0$