Let $f$ be a differentiable function in $\left(0,\frac{\pi }{2}\right)$. If ${\int }_{\cos x}^1{t}^{2}f\left(t\right)dt={\sin }^{3}x+\cos x$, then $\frac{1}{\sqrt{3}}{f}^{\text{'}}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

${\int }_{\cos x}^1{t}^{2}f\left(t\right)dt={\sin }^{3}x+\cos x$

On differentiating, we get

$f\left(\cos x\right)\sin x·{\cos }^{2}x=3{\sin }^{2}x\cos x-\sin x$

$f\left(\cos x\right)=3\tan x-{\sec }^{2}x$

Again differentiating, we get

$-\sin x{f}^{\text{'}}\left(\cos x\right)=3{\sec }^{2}x-2{\sec }^{2}x\tan x$

When $\cos x=\frac{1}{\sqrt{3}}$ then $\sec x=\sqrt{3}$, $\tan x=\sqrt{2} \& \sin x=\frac{\sqrt{2}}{\sqrt{3}}$

Then $-\frac{\sqrt{2}}{\sqrt{3}}{f}^{\text{'}}\left(\frac{1}{\sqrt{3}}\right)=3\times 3-2\times 3\sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{3}}{f}^{\text{'}}\left(\frac{1}{\sqrt{3}}\right)=6-\frac{9}{\sqrt{2}}$