$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $.  then find the minimum value of $x^3 y + xy^3 +4$.
I assume  $ x = r \sin (w)$ and $ y = r\cos(w) $. $  x^3 y + xy^3 +4 = L $  which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my  Humble request to help me after that.

Your method of solution is good. You just need to continue. By setting $x=r\cos\theta$ and $y=r\sin\theta$, we get$$\cos\theta\sin\theta=\frac1{r^2}$$ and so $\tfrac23\leqslant r^2\leqslant2$, as you found. We have $$x^3y+xy^3+4=r^2-r^4+4=\tfrac{17}4-(r^2-\tfrac12)^2,$$which is minimum when $|r^2-\frac12|$ is maximum, namely when $r^2=2$.
