Let $$y = f(x) = \sin^3\left(\frac{\pi}{3} \cos\left(\frac{\pi}{3\sqrt{2}}(-4x^3 + 5x^2 + 1)^{\frac{3}{2}}\right)\right)$$. Then, at $$x = 1$$,

To find the relation between $$y$$ and $$y'$$ at $$x = 1$$, we first compute the value of the function $$y = f(1)$$ and then find its derivative $$y' = f'(1)$$ using the chain rule.

The given function is:

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

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Step 1: Evaluate $$y$$ at $$x = 1$$

Let $$u(x) = -4x^3 + 5x^2 + 1$$. At $$x = 1$$:

$$u(1) = -4(1)^3 + 5(1)^2 + 1 = 2$$

Now, evaluate the argument of the cosine function, let it be $$\theta(1)$$:

$$\theta(1) = \frac{\pi}{3\sqrt{2}} (u(1))^{\frac{3}{2}} = \frac{\pi}{3\sqrt{2}} (2)^{\frac{3}{2}} = \frac{\pi}{3\sqrt{2}} (2\sqrt{2}) = \frac{2\pi}{3}$$

Next, find the value inside the sine function:

$$\cos(\theta(1)) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$

$$\text{Argument of sine} = \frac{\pi}{3} \cos\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} \left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$

Finally, evaluate the function $$y$$ at $$x = 1$$:

$$y(1) = \sin^3\left(-\frac{\pi}{6}\right) = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$$

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Step 2: Differentiate $$y$$ with respect to $$x$$ using the chain rule

Let $$y = \sin^3(v)$$, where $$v = \frac{\pi}{3}\cos(\theta)$$ and $$\theta = \frac{\pi}{3\sqrt{2}}(u(x))^{\frac{3}{2}}$$.

$$y' = 3\sin^2(v) \cdot \cos(v) \cdot v'$$

$$v' = -\frac{\pi}{3}\sin(\theta) \cdot \theta'$$

$$\theta' = \frac{\pi}{3\sqrt{2}} \cdot \frac{3}{2}(u(x))^{\frac{1}{2}} \cdot u'(x)$$

Differentiating $$u(x) = -4x^3 + 5x^2 + 1$$ gives:

$$u'(x) = -12x^2 + 10x$$

Evaluating the component derivatives at $$x = 1$$:

$$u'(1) = -12(1)^2 + 10(1) = -2$$

$$\theta'(1) = \frac{\pi}{3\sqrt{2}} \cdot \frac{3}{2}\sqrt{2} \cdot (-2) = -\pi$$

Now substitute $$\theta'(1)$$ into the expression for $$v'(1)$$, noting that $$\theta(1) = \frac{2\pi}{3}$$:

$$v'(1) = -\frac{\pi}{3}\sin\left(\frac{2\pi}{3}\right) \cdot (-\pi) = \frac{\pi^2}{3} \cdot \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi^2}{2\sqrt{3}}$$

Now compute $$y'(1)$$, using $$v(1) = -\frac{\pi}{6}$$:

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

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

$$y'(1) = 3 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\pi^2}{2\sqrt{3}} = \frac{3\pi^2}{16}$$

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Step 3: Establish the final relationship

We have the values:

$$y = -\frac{1}{8} \implies 8y = -1 \implies 16y = -2$$

$$y' = \frac{3\pi^2}{16} \implies 16y' = 3\pi^2$$

Multiplying $$y'$$ by 2:

$$2y' = 2\left(\frac{3\pi^2}{16}\right) = \frac{3\pi^2}{8} = -3\pi^2\left(-\frac{1}{8}\right) = -3\pi^2 y$$

Rearranging the terms:

$$2y' + 3\pi^2 y = 0$$

Therefore, the correct relation corresponds to statement (2).