Given $x, y in mathbb{R}^+$ and $ x^3 + y^3 = x-y$. Prove $x^2 + 4y^2 < 1$

For $x, y in mathbb{R}^+$ , I am given that $ x^3 + y^3 = x-y$. I have to prove that $x^2 + 4y^2 < 1$. Now I have $ y + y^3 = x(1-x^2) $. Since $y + y^3 > 0 $ and $ x > 0$, we have $1 – x^2 > 0$. Which means that $ 0 < x < 1$. Also, since $x^3 + y^3 > 0$, we have $x – y > 0$. So, $x > y$. So, we get $0 < y < x < 1$. Using AM-GM inequality, I also get

$$ (x^3 + y^3 + 1) geqslant 3xy $$

What else can be dome here ?

Thanks