Let $n=1$, $f(t)=t^2 + |t|^{7/2}\sin(1/|t|)$ for $t\ne0$, $f(0):=0$. Then $f'(0)=0$ and $f''(0)=2>0$, so that $0$ is a strict local minimum of $f$. However, $f''(t)\sim-|t|^{-1/2}\sin(1/|t|)$ as $t\to0$, and so, $f$ is not convex (let alone strictly convex) in any neighborhood of $0$.

Let $n=1$, $f(t)=t^2 + |t|^{7/2}\sin(1/|t|)$ for $t\ne0$, $f(0):=0$. Then $f'(0)=0$ and $f''(0)=2>0$, so that $0$ is a strict local minimum of $f$. However, $f''(t)\sim-|t|^{-1/2}\sin(1/|t|)$ as $t\to0$, and so, $f$ is not convex (let alone strictly convex) in any neighborhood of $0$.

Here are the graphs $\{(t,f(t))\colon|t|<0.1\}$ (left) and $\{(t,f''(t))\colon|t|<0.1\}$ (right).