jmc

Clearly, the maximum occurs when $x$ is positive and $y$ is negative. Let $z=-y,$ so $z$ is positive, and $y=-z.$ Then $$\frac{x-y}{x^4+y^4+6}=\frac{x+z}{x^4+z^4+6}.$$By AM-GM, $$x^4+1+1+1\ge 4\sqrt[4]{x^4}=4x,$$and $$z^4+1+1+1\ge 4\sqrt[4]{z^4}=4z.$$Then $x^4+z^4+6\ge 4(x+z),$ which implies $$\frac{x+z}{x^4+z^4+6}\le \frac{1}{4}.$$Equality occurs when $x=z=1,$ so the maximum value is $\boxed{\frac{1}{4}}.$