If $\int \frac{1}{1-\sin ^4 x} d x=\frac{1}{2} \tan x+A \tan ^{-1}\{f(x)\}+C$, then

$A=\frac{1}{2 \sqrt{2}}$ and $f(x)=\sqrt{2}$ tan $x$

Let

$I=\int \frac{1}{1-\sin ^4 x} d x=\int \frac{1}{\left(1-\sin ^2 x\right)\left(1+\sin ^2 x\right)} d x$

$\Rightarrow I=\int \frac{\sec ^2 x}{1+\sin ^2 x} d x=\int \frac{1+\tan ^2 x}{1+2 \tan ^2 x} d(\tan x)$

$ \Rightarrow I=\frac{1}{2} \int \frac{\tan ^2 x+\frac{1}{2}+\frac{1}{2}}{\tan ^2 x+\frac{1}{2}} d(\tan x)$

$\Rightarrow I=\frac{1}{2} \int\left(1+\frac{1}{2} . \frac{1}{\tan ^2 x+\frac{1}{2}}\right) d(\tan x)$

$\Rightarrow I=\frac{1}{2}\left[\tan x+\frac{\sqrt{2}}{2} \tan ^{-1}(\sqrt{2} \tan x)\right]+C$

$\Rightarrow I=\frac{1}{2} \tan x+\frac{\sqrt{2}}{4} \tan ^{-1}(\sqrt{2} \tan x)+C$

Hence, $A=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}}$ and $f(x)=\sqrt{2} \tan x$