Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x,y) = (y-x^2)(y-2x^2)$, I must prove that $(0,0)$ is a minimal point for all lines along the origin, but it's not a minimal point for $f$. This question is similar to this one: Does having a minimum along all lines imply that the function has a minimum?.
I notice that we have two sets, our function is $0$ in $$U_1 = \{(x,y) \in \mathbb{R}^2: y=x^2 \ \text{or} \ y=2x^2\}$$ and $f$ is negative in $$U_2 = \{(x,y) \in \mathbb{R}^2: x^2<y<2x^2\}.$$
It looks clear that $(0,0)$ isn't a minimum point for $f$. So i must prove that the origin is a minimum for all lines. My attempt was to consider the following function: $$g(x) := f(x,\lambda x), \lambda \in \mathbb{R}$$ so we can do $g'(x) = 8x^3 - 9\lambda x^2 +2\lambda^2 x$. This give us $g'(x)=0$ when $x=0$ and we can show that $(0,0)$ is a minimum (I know that we have 2 other roots). My question is: there are easier ways? Some elegant proof maybe.