\sin^2 x+\cos^2 =1

\sin^2 x+\cos^2x +2\sin x\cos x=1+2\sin x\cos x

\sin^4 x+\cos^4 x-\sin x\cos x

= \sin^4 x+\cos^4+2\sin^2 x\cos^2 x-2\sin^2 x\cos^2 x-\sin x\cos x

=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x-\sin x\cos x

=1- 2\sin^2 x\cos^2 x-\sin x\cos x

=1- \dfrac{1}{2} \sin^2 2x - \dfrac{1}{2} \sin 2x

Let t = \sin 2x, t\in [-1,1]

f(x) = -\dfrac{1}{2}t^2-\dfrac{1}{2}t+1

=-\dfrac{1}{2}(t+\dfrac{1}{2})^2 +\dfrac{9}{8}

When t = -\dfrac{1}{2} , the maximum of f(x) is \dfrac{9}{8}

When t = 1, the minimum of f(x) is

-\dfrac{1}{2}\cdotp \dfrac{9}{4}+\dfrac{9}{8} = 0

Therefore, 0 \leq f(x) \leq \dfrac{9}{8}