There is an exercise in my lists about those functions: $f(x, y) = (y-3x^2)(y-x^2) = 3 x^4-4 x^2 y+y^2$
$g(t) = f(vt) = f(at, bt); a, b \in \mathbf{R}$
It asks me to prove that $t = 0$ is a local minimum of $g$ for all $a, b \in \mathbf{R}$
I did it easily: $g(t) = 3 a^4 t^4-4 a^2 t^2 b t+b^2 t^2$ g'(t) = 2 b^2 t-12 a^2 b t^2+12 a^4 t^3 g''(t) = 2 b^2-24 a^2 b t+36 a^4 t^2
It is a critical point: g'(0) = 0; \forall a, b
Its increasing for all a, b: g''(0) = 2b^2 > 0; \forall b \ne 0 and $b = 0 \implies g(t) = 3 a^4 t^4$ which has only one minimum, at $0$, and no maximum
However, it also asks me to prove that $(0, 0)$ is not a local minimum of $f$. How can this be possible? I mean, if $(0, 0)$ is a minimum over every straight line that passes through it, then, in this point, $f$ should be increasing in all directions, no?