If $f\left(x\right)=-\left({\sin }^{2}x+{\cos }^{5}x\right)...................\left(i\right)$

 To find:$\underset{x\to 0}{\lim }\frac{1}{x}{f}^{\text{'}}\left(x\right)$

Differentiating equation $\left(i\right)$ with respect to $x$, we get

$f^{\text{'}}\left(x\right)=-\left[2\sin x.\cos x+5{\cos }^{4}x.\left(-\sin x\right)\right]$

$f^{\text{'}}\left(x\right)=-\sin x\left[2.\cos x-5{\cos }^{4}x\right]$

Now,

 $\underset{x\to 0}{\lim }\frac{1}{x}{f}^{\text{'}}\left(x\right)$

$=\underset{x\to 0}{\lim }\frac{1}{x}\times \left(-\sin x\right)\left[2\cos x-5{\cos }^{4}x\right]$

$=\underset{x\to 0}{\lim }\frac{-\sin x}{x}\times \underset{x\to 0}{\lim } \left[2\cos x-5{\cos }^{4}x\right]$

$=-1\times \left(2-5\right)$

$=3$

Hence, option $3$ is correct.