If $\int \frac{\mathrm{d}x}{{\sin }^{4}x+{\cos }^{4}x}=\frac{1}{\sqrt{2}}{\tan }^{-1}f\left(x\right)+C$ then

$\text{ Dividing the numerator and denominator by }$

${\cos }^{4}x\text{, the given integral can be written as }$

$\begin{array}{l}\int \frac{{\sec }^{2}x\left(1+{\tan }^{2}x\right)}{{\tan }^{4}x+1}\mathrm{d}x=\int \frac{1+t^2}{1+t^4}\mathrm{d}t(t=\tan x)\\ =\int \frac{1+1/t^2}{t^2+1/t^2}\mathrm{d}t=\int \frac{1+1/t^2}{{\left(t-\frac{1}{t}\right)}^2+2}\mathrm{d}t\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}\left(t-\frac{1}{t}\right)+C\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\tan x-\cot x}{\sqrt{2}}+C\end{array}$

Thus $f\left(x\right)=\frac{\tan x-\cot x}{\sqrt{2}}\text{ so }f\left(\frac{\pi }{4}\right)=0$