Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ with $f(x,y)=x^2+y^2-2x^2y - \frac{4x^6y^2}{(x^4+y^2)^2}$ for $(x,y)\neq(0,0)$ and $f(0,0)=0$. Show that the function $t\mapsto f(tv)$ has a minimum in $t=0$ for every direction $v\in\mathbb{R}^2$, but $f(0,0)$ is not a local minimum of $f$.
I'm new to this and trying to figure out how to best proceed here. Do I understand correctly that I must first find the second derivative of the function $g(t)=f(vt)=f(at,bt)$ with $a,b\in\mathbb{R}$, e.g.
$$g(t)=a^2t^2 + b^2t^2 - 2a^2t^2bt - \frac{4a^6t^6b^2t^2}{(a^4t^4+b^2t^2)^2}$$
and then set $t=0$ and see if $\nabla^2g(t)\geq0$ to then use the first derivative and see if $g(0)=0$?
If that is correct, how would I show that $f(0,0)$ is not a local mimum of $f$?
HINT for the second part: Consider what happens on the curve $y=x^2$.