\(x^3+y^3=x-y\)

\(\Rightarrow x^3+y^3-x+y=0\)

\(\Rightarrow x^3-x+y^3+y=0\)

\(\Rightarrow x\left(x^2-1\right)+y\left(y^2+1\right)=0\)

But \(x,y\ge0\)

\(\Rightarrow x^2,y^2\ge0\Leftrightarrow x\left(x^2-1\right);y\left(y^2+1\right)\ge0\)

=> \(x\left(x^2-1\right)andy\left(y^2+1\right)\) aren't two numbers opposite

\(\Rightarrow x\left(x^2-1\right)+y\left(y^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x\left(x^2-1\right)=0\\y\left(y^2+1\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\\\left[{}\begin{matrix}y=0\\y^2+1=0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\\\left[{}\begin{matrix}y=0\\y\in\phi\end{matrix}\right.\end{matrix}\right.\)

But \(x\ge0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\Rightarrow MaxA=x^2+y^2=1^2+0^2=1+0=1\)

You were wrong. \(x^2-1\ge0\) is not always true

@Phan Thanh Tinh, in the subject said \(x,y\ge0\), so I only apply :]

Dao Trong Luan, the statement \(x\left(x^2-1\right)\ge0\) is wrong when x < 1