1. Given equation

We have

$$\int_{\cos x}^{1} t^2 f(t)\,dt = \sin^3 x + \cos x, \qquad x\in\left(0,\frac{\pi}{2}\right).$$

We need to find

$$\frac{1}{\sqrt{3}}f'\left(\frac{1}{\sqrt{3}}\right).$$

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2. Differentiate both sides with respect to $x$

Using Leibniz rule,

$$\frac{d}{dx}\left(\int_{\cos x}^{1} t^2 f(t)\,dt\right) = -\, (\cos^2 x) f(\cos x)\cdot \frac{d}{dx}(\cos x).$$

Since $\frac{d}{dx}(\cos x)=-\sin x$,

$$\frac{d}{dx}\left(\int_{\cos x}^{1} t^2 f(t)\,dt\right) = \cos^2 x\, f(\cos x)\sin x.$$

Now differentiate the RHS:

$$\frac{d}{dx}(\sin^3 x + \cos x)=3\sin^2 x\cos x-\sin x.$$

Hence,

$$\cos^2 x\, f(\cos x)\sin x = 3\sin^2 x\cos x-\sin x.$$

Since $x\in(0,\pi/2)$, we have $\sin x>0$, so divide by $\sin x$:

$$\cos^2 x\, f(\cos x)=3\sin x\cos x-1.$$

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3. Express in terms of $t=\cos x$

Let

$$t=\cos x, \qquad t\in(0,1).$$

Then

$$\sin x=\sqrt{1-t^2}.$$

So

$$t^2 f(t)=3t\sqrt{1-t^2}-1.$$

Therefore,

$$f(t)=\frac{3t\sqrt{1-t^2}-1}{t^2}.$$

Rewrite as

$$f(t)=\frac{3\sqrt{1-t^2}}{t}-\frac{1}{t^2}.$$

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4. Differentiate $f(t)$

We compute

$$f(t)=3(1-t^2)^{1/2} t^{-1} - t^{-2}.$$

Differentiate term by term.

For

$$g(t)=\frac{\sqrt{1-t^2}}{t},$$

using product rule:

$$g(t)=(1-t^2)^{1/2} t^{-1}.$$

Then

$$g'(t)=\left(-\frac{t}{\sqrt{1-t^2}}\right)t^{-1}+(1-t^2)^{1/2}(-t^{-2}).$$

So

$$g'(t)=-\frac{1}{\sqrt{1-t^2}}-\frac{\sqrt{1-t^2}}{t^2}.$$

Thus,

$$f'(t)=3g'(t)+2t^{-3}$$

that is,

$$f'(t)= -\frac{3}{\sqrt{1-t^2}}-\frac{3\sqrt{1-t^2}}{t^2}+\frac{2}{t^3}.$$

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5. Evaluate at $t=\frac{1}{\sqrt{3}}$

First,

$$t=\frac{1}{\sqrt{3}}, \qquad t^2=\frac13,$$

so

$$1-t^2=1-\frac13=\frac23,$$

therefore

$$\sqrt{1-t^2}=\sqrt{\frac23}=\frac{\sqrt2}{\sqrt3}.$$

Now evaluate each term:

$$-\frac{3}{\sqrt{1-t^2}}=-\frac{3}{\sqrt{2/3}}=-\frac{3\sqrt3}{\sqrt2},$$

and

$$-\frac{3\sqrt{1-t^2}}{t^2}=-3\cdot \frac{\sqrt2}{\sqrt3}\cdot 3=-3\sqrt6,$$

and

$$\frac{2}{t^3}=2\left(\sqrt3\right)^3=6\sqrt3.$$

So

$$f'\left(\frac{1}{\sqrt3}\right)= -\frac{3\sqrt3}{\sqrt2}-3\sqrt6+6\sqrt3.$$

Now multiply by $\frac{1}{\sqrt3}$:

$$\frac{1}{\sqrt3}f'\left(\frac{1}{\sqrt3}\right) = -\frac{3}{\sqrt2}-3\sqrt2+6.$$

Since

$$3\sqrt2=\frac{6}{\sqrt2},$$

we get

$$-\frac{3}{\sqrt2}-3\sqrt2 = -\frac{3}{\sqrt2}-\frac{6}{\sqrt2}=-\frac{9}{\sqrt2}.$$

Hence,

$$\frac{1}{\sqrt3}f'\left(\frac{1}{\sqrt3}\right)=6-\frac{9}{\sqrt2}.$$

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6. Compare with options

The value is

$$6-\frac{9}{\sqrt2}.$$

This matches Option B.