y=\sin x-\cos x+\sin x\cos x

=\sin x-\cos x -\dfrac{(\sin x-\cos x)^2-1}{2}

Let t=\sin x-\cos x

=\sqrt{2}(\dfrac{\sqrt{2} }{2}\sin x-\dfrac{\sqrt{2} }{2} \cos x)

=\sqrt{2}\sin(x-\dfrac{\pi}{4})

-\sqrt{2} < t <\sqrt{2}

Then,

y=-\dfrac{1}{2} t^2+t+\dfrac{1}{2}

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

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

Therefore,

When t=-1, t_{max}=1