solution: 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 $\boxed{\frac{1}{4}}.$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Find the maximum value of
  \[\frac{x - y}{x^4 + y^4 + 6}\]over all real numbers $x$ and $y.$
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: \frac{1}{4}
extra_info.question: Find the maximum value of
  \[\frac{x - y}{x^4 + y^4 + 6}\]over all real numbers $x$ and $y.$
extra_info.split: test