因为 $y=\dfrac{\sin x\cdot\cos x-(\sin x+\cos x)+1}{2+2\sin x\cdot\cos x}$，令 $t=\sin x+\cos x=\sqrt{2}\sin(x+\dfrac{\pi}{4})\in[-\sqrt{2},\sqrt{2}]$，则 $\sin x\cos x=\dfrac{1}{2}(1-t^2)$，于是 $y=\dfrac{\dfrac{1}{2}(t^2-1)-t+1}{2+(t^2-1)}=\dfrac{1}{2}-\dfrac{t}{t^2+1}$．令 $g(t)=\dfrac{t}{t^2+1}(t\in[-\sqrt{2},\sqrt{2}])$，则 $g^\prime(t)=\dfrac{1-t^2}{(t^2+1)^2}$．由 $g^\prime(t)=0$ 知 $t=-1$ 或 $1$．
因为 $g(-\sqrt{2})=-\dfrac{\sqrt{2}}{3},g(-1)=-\dfrac{1}{2},g(1)=\dfrac{1}{2},g(\sqrt{2})=\dfrac{\sqrt{2}}{3}$，于是 $g(t)$ 的最小值是 $g(-1)=-\dfrac{1}{2}$，所以 $y$ 的最大值是 $\dfrac{1}{2}-(-\dfrac{1}{2})=1$．