Here is my counterexample:
Take $f(x)=\|x\|^4$, then for any $x\neq 0$, by spherical symetry, $x$ is a eigenvector of $Hf|_x$.
We can choose $x=v$ an eigenvector on $\frac{1}{2}(A+A^T)$ with negatif eigenvalue and study then $Hg|_{\lambda x}$ for $\lambda\in [0,\infty[$. By continuity there exists $\lambda$ such that $Hg|_{\lambda x}(x)=0$