The function $\mathrm{f}(x)=\sin ^4 x+\cos ^4 x$ increases if

Function simplification:
Begin by expressing $\sin^4 x + \cos^4 x$ as $(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.
Using the identity $2\sin x \cos x = \sin(2x)$, we write $2\sin^2 x \cos^2 x = \frac{1}{2}\sin^2(2x)$, yielding $f(x) = 1 - \frac{1}{2}\sin^2(2x)$.

Differentiation:
The derivative is $f'(x) = -\frac{1}{2} \cdot 2\sin(2x) \cdot 2\cos(2x) = -2\sin(2x)\cos(2x)$.
Apply the identity $2\sin(2x)\cos(2x) = \sin(4x)$ to obtain $f'(x) = -\sin(4x)$.

Increasing condition:
The function increases where $f'(x) > 0$, which requires $-\sin(4x) > 0$ or $\sin(4x) < 0$.
The sine function is negative when its argument is in the interval $(2n\pi + \pi, 2n\pi + 2\pi)$ for integer $n$.
Substitute $\theta = 4x$ to get $2n\pi + \pi < 4x < 2n\pi + 2\pi$, and divide by 4:
$(2n+1)\frac{\pi}{4} < x < (n+1)\frac{\pi}{2}$.

Interval evaluation:
For $n=0$, the interval is $\frac{\pi}{4} < x < \frac{\pi}{2}$, where $\sin(4x) < 0$ and $f'(x) > 0$.
This corresponds to option B. Other intervals either yield $f'(x) \leq 0$ or do not maintain a consistent sign for $f'(x)$.

The function increases on $\frac{\pi}{4} < x < \frac{\pi}{2}$, so the correct choice is $\boxed{B}$.