If $f\left(x\right)=\frac{{\cos }^{2}x}{1+{\sin }^{2}x}$, then $f\left(\frac{\pi }{4}\right)-3f^{\text{'}}\left(\frac{\pi }{4}\right)=$

$f\left(x\right)=\frac{{\cos }^{2}x}{1+{\sin }^{2}x}\Rightarrow f\left(\frac{\pi }{4}\right)=\frac{{\cos }^2\left(\frac{\pi }{4}\right)}{1+{\sin }^2\left(\frac{\pi }{4}\right)}=\frac{1}{3} ...\left(1\right)$

$\Rightarrow \ln \left(f\left(x\right)\right)=\ln \left(\frac{{\cos }^{2}x}{1+{\sin }^{2}x}\right)=2\ln \left|\cos x\right|-\ln \left|1+{\sin }^{2}x\right|$

Differentiating both sides w.r.t. $x$,

$\Rightarrow \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=\frac{-2\sin x}{\cos x}-\frac{2\sin x\cos x}{1+{\sin }^{2}x}=-2\tan x-\frac{\sin 2x}{1+{\sin }^{2}x}$

Putting $x=\frac{\pi }{4}$,

$\Rightarrow \frac{f^{\text{'}}\left({\displaystyle \frac{\pi }{4}}\right)}{f\left(\frac{\pi }{4}\right)}=-2-\frac{1}{1+{\displaystyle \frac{1}{2}}}=\frac{-8}{3}$

$\Rightarrow f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{-8}{3}\times \frac{1}{3}=\frac{-8}{9}$

So, $f\left(\frac{\pi }{4}\right)-3f^{\text{'}}\left(\frac{\pi }{4}\right)=\frac{1}{3}+3\times \frac{8}{9}=3$