+ \(x^2+y^2-xy=4\) \(\Rightarrow2x^2+2y^2-2xy=8\)
\(\Rightarrow x^2+y^2+\left(x-y\right)^2=8\)
\(\Rightarrow x^2+y^2\le8\) ( do \(\left(x-y\right)^2\ge0\forall x,y\) )
Max \(x^2+y^2=8\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2+y^2-xy=4\end{matrix}\right.\Leftrightarrow x=y=2\)