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

For a function to be monotonically increasing, $f^{\prime }(x)>0$.Given $f(x)={\sin }^{4}x+{\cos }^{4}x$.Differentiating with respect to $x$:
$$f^{\prime }(x)=4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x$$
$$f^{\prime }(x)=4\sin x\cos x({\sin }^{2}x-{\cos }^{2}x)$$
Using double angle formulas $\sin 2x=2\sin x\cos x$ and $\cos 2x={\cos }^{2}x-{\sin }^{2}x$:
$$f^{\prime }(x)=2(2\sin x\cos x)(-({\cos }^{2}x-{\sin }^{2}x))$$
$$f^{\prime }(x)=2\sin 2x(-\cos 2x)=-\sin 4x$$
For $f(x)$ to increase, $f^{\prime }(x)>0⟹-\sin 4x>0⟹\sin 4x<0$.$\sin \theta <0$ when $\pi <\theta <2\pi$.Substituting $\theta =4x$:
$$\pi <4x<2\pi ⟹\frac{\pi }{4}<x<\frac{\pi }{2}$$
In this interval, the function is monotonically increasing.