Let $$\begin{aligned} f(x,y) & = 2 x^6 - x^4 y^2 - x^2 y^4 + 2 y^6 \\ & = x^6 + y^6 + (x^2 - y^2)^2 (x^2 + y^2) . \end{aligned}$$ Then $f$ is a strictly positive (except at the origin, of course) homogeneous polynomial of degree $6$, and hence $d^j f = 0$ for $j < 6$ and $d^6 f > 0$. On the other hand, $$\partial_{xx} f(0,y) = -2 y^4 < 0$$ whenever $y \ne 0$, and so $f$ is not convex near $0$.

Let $$\begin{aligned} f(x,y) & = x^4 - x^2 y^2 + y^4 \\ & = \tfrac{1}{2} x^4 + \tfrac{1}{2} y^4 + \tfrac{1}{2} (x^2 - y^2)^2 . \end{aligned}$$ Then $f$ is a strictly positive (except at the origin, of course) homogeneous polynomial of degree $4$, and hence $d^j f(\vec 0) = 0$ for $j < 4$ and $d^4 f(\vec 0) > 0$ (indeed: $d^4 f(\vec 0)(\vec h, \vec h, \vec h, \vec h) = 4! f(\vec h) > 0$ whenever $\vec h \ne \vec 0$). On the other hand, $$\partial_{xx} f(0,y) = -2 y^2 < 0$$ whenever $y \ne 0$, and so $f$ is not convex near $0$.