$x^2 + y^2+xy = 1$ , 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.

Lagrangian multiplier for constrained function 
$$L(x,y)= F(x,y)-\lambda G(x,y) $$
set in the order given, partially differentiating with respect to both variables
$$ \dfrac{F_x}{F_y}=\dfrac{G_x}{G_y}= \lambda $$
$$\dfrac{2x+y}{2y+x}=\dfrac{3x^2y+x^3}{x^3+3y^2} $$
Cross multiplying to simplify we get two conditions
$$ (x^2-y^2)(x^2+y^2-xy)=0 $$
Minimal value verifiable by sign of second partial derivatives of $L(x,y)$..
