jmc

We can write $2(x^2+y^2)=x+y$ as $2x^2+2y^2=x+y.$ Then $2x^2+4xy+2y^2=x+y+4xy,$ so $$4xy=2(x^2+2xy+y^2)-(x+y)=2(x+y{)}^2-(x+y).$$Also, $$\begin{array}{rl}(x-y{)}^2& =x^2-2xy+y^2\\ & =(x+y{)}^2-4xy\\ & =(x+y)-(x+y{)}^2.\end{array}$$Completing the square in $x+y,$ we get $$(x-y{)}^2=\frac{1}{4}-{\left(x+y-\frac{1}{2}\right)}^2\le \frac{1}{4},$$so $x-y\le \frac{1}{2}.$

Equality occurs when $x=\frac{1}{2}$ and $y=0,$ so the maximum value is $\boxed{\frac{1}{2}}.$