Find largest ellipse in basin of attraction

Consider the following system: $$ \begin{align} x' &= -x+y^2 \\ y' &= -2y + 3x^2 \\ \end{align} $$ I am asked to find the largest ellipse of the form: $$ V(x,y) = \tfrac{x^2}{2} + \tfrac{y^2}{4} = r$$ contained in the region of attraction of the system.

We can use $V(x,y)$ as a Lyapunov function of the system and we get: $$V'(x,y) = xx' + \tfrac{y}{2}y' = -x^2 +xy^2 -y^2 + \tfrac{3}{2} x^2y$$

I deduced we are looking for $r$ such that: \begin{align} r &= \min_{(x,y) \neq (0,0)} \tfrac{x^2}{2} + \tfrac{y^2}{4}\\ &\text{ s.t } -x^2 +xy^2 -y^2 + \tfrac{3}{2} x^2y=0 \end{align} I have not been able to solve this problem analytically using the Lagrangian (a 3rd-degree polynomial comes up). Wolfram Alpha solution seems to be right ($r \simeq 0.4782$) when plotted though:

Is this way of reasoning sound? Is there some simplification that allows us to get an analytical solution? Thanks in advance

EDIT: I have found the answer in the book, apparently the correct result is $\frac{1}{9}$. So there must be an error in my reasoning I cannot find. The book offers no solution, so I still don't know how to correctly solve the exercise.

Another way to solve

$$ \min_r \ \ \text{s. t.}\ \ \cases{-x^2 + x y^2 - y^2 + 3/2 x^2 y = 0\\ \frac{x^2}{2}+\frac{y^2}{4}=r} $$

is by way of tangency. After substitution of $x$ and posterior squaring we have

$$ \frac{144 r^2 y^2}{17}-\frac{192 r^2 y}{17}+\frac{64 r^2}{17}-\frac{104 r y^4}{17}+\frac{32 r y^2}{17}+y^6+\frac{12 y^5}{17}+\frac{4 y^4}{17}=0 $$

This polynomial should have at least a double root (tangengy) so

$$ \frac{144 r^2 y^2}{17}-\frac{192 r^2 y}{17}+\frac{64 r^2}{17}-\frac{104 r y^4}{17}+\frac{32 r y^2}{17}+y^6+\frac{12 y^5}{17}+\frac{4 y^4}{17}=(y-y_1)^2(y^4+a y^3+b y^2+c y +d) $$

and the coefficients which guarantee nullity are

$$ \left\{ \begin{array}{l} d y_1^2-\frac{64 r^2}{17} = 0\\ c y_1^2-2 d y_1+\frac{192 r^2}{17} = 0\\ b y_1^2-2 c y_1+d-\frac{144 r^2}{17}-\frac{32 r}{17} = 0\\ a y_1^2-2 b y_1+c = 0\\ y_1^2 -2 a y_1+b+\frac{104 r}{17}-\frac{4}{17} = 0\\ a-2 y_1-\frac{12}{17}=0 \\ \end{array} \right. $$

which gives the possible real solutions

$$ r = \cases{ 0\\ 0.4782377831432382\\ 3.5153505723608345\\ 12.75203735221895 } $$