Let y = f x = sin 3 π 3 cos π 3 2 - 4 x 3 + 5 x 2 + 1 3 2 . Then, at x = 1 ,

The given equation can be written as

$y={\sin }^3\left(\frac{\pi }{3}\cos g\left(x\right)\right)$ 

where,

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

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

$\Rightarrow g^{\text{'}}\left(1\right)=\frac{\pi \sqrt{2}}{2\sqrt{2}}\left(-2\right)=-\mathrm{\pi }$

And,

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

Now,

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

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

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

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

Therefore,

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