 Solution

Near a nonzero stationary point $\mathbf{p}$, Taylor expansion of the gradient gives $\nabla h(\mathbf{p}+\mathbf{v})=H(\mathbf{p})\mathbf{v}+O(|\mathbf{v}{|}^2)$. Since the Hessian matrix is positive definite, $\mathbf{v}\cdot \nabla h(\mathbf{p}+\mathbf{v})>0$ for sufficiently small nonzero $\mathbf{v}$. Thus the gradient arrows point away from each of the four local minima, with different strengths in the two eigenvector directions.

At the origin, on the $x$-axis the gradient is $(6x^5,0)$ and on the $y$-axis it is $(0,6y^5)$, so arrows point outward. On $y=x$ the gradient is $6x^3(x^2-8)(1,1)$, and on $y=-x$ it is $6x^3(x^2-8)(1,-1)$, so arrows point toward the origin for small $|x|$. Away from the axes the leading terms are $(-48x{y}^2,-48x^{2}y)$, pointing inward in both coordinate signs. The narrow outward sectors close to the axes are visible in the exact vector field.