Using the method of Lagrange multipliers, find the extreme values of a function Using the method of Lagrange multipliers, find the extreme values of the function
$f(x,y)= \frac{2y^3}{3} + 2x^2 +1$ on the ellipse $5x^2 + y^2 = 1/9$
 . Identify the (absolute) maximal and minimal values of f
taken on the ellipse.
currently I have that $\nabla f(x,y)=L*\nabla g(x,y)$ where $g(x,y)=5x^2 + y^2-1/9$
this leads to: $4x-10Lx=0$, so $x=0$ or $L=2/5$
and:$2y^2-2Ly=0$, so $y=0$ or $y=L$.
where do I go from here? andy help would be greatly appreciated, thank you in advance.

$\dfrac{\partial f}{\partial x} = 4x= \gamma\cdot 10x$, and $\dfrac{\partial f}{\partial y} = 2y^2 = \gamma\cdot 2y$. So: $2x(2 - 5\gamma) = 0$, and $2y(y - \gamma) = 0$. So 
case 1: $x = 0$, so $y^2 = \dfrac{1}{9}$. So $y = \dfrac{1}{3}$ or $-\dfrac{1}{3}$. So:
$f(0,\frac{1}{3}) = \dfrac{83}{81}$, and $f(0,-\frac{1}{3}) = \dfrac{79}{81}$
case 2: $x \neq 0$, and $\gamma = \dfrac{2}{5}$, and $y = 0$ or $y = \dfrac{2}{5}$. If $y = 0$, then $x = \dfrac{1}{3\sqrt{5}}$ or $x = -\dfrac{1}{3\sqrt{5}}$, and in both cases $f(\frac{1}{3\sqrt{5}}, 0) = \dfrac{47}{45}$. If $y = \dfrac{2}{5}$, then $5x^2 = -\dfrac{12}{225} < 0$, and there is no solution in this case.
So: $f_{min} = \dfrac{79}{81} = 0.975$, and $f_{max} = \dfrac{47}{45} = 1.044$
