Just write cos(x-y)=-(x^2+1)/2x and put the condition, that this must be between -1 and 1

Rewrite as $(x+\cos(x-y))^2+\sin^2(x-y)=0$ hence $x+\cos(x-y)=0, \sin(x-y)=0$. Then $\cos(x-y)=\pm 1 \implies x=\pm 1$. Checking the solutions gives $x=1,y\equiv \pi+1 \mod 2\pi \vee x=-1, y\equiv -1 \mod 2\pi$.