By the definition of differentiability, for each $n$ there exists $\tilde{\theta}_n$ lying on the segment $[\theta_0, \hat{\theta}_n]$ such that \begin{align*} \sqrt{n}\bigl(\Phi(\hat{\theta}_n) - \Phi(\theta_0)\bigr) = \nabla_\theta \Phi(\tilde{\theta}_n)^\top \sqrt{n}(\hat{\theta}_n - \theta_0). \end{align*} Since $\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} Z$, it is bounded in probability, which gives $\|\hat{\theta}_n - \theta_0\| = O_{\mathbb{P}}(1/\sqrt{n})$. Since $\tilde{\theta}_n$ lies between $\theta_0$ and $\hat{\theta}_n$, we also have $\|\tilde{\theta}_n - \theta_0\| \leq \|\hat{\theta}_n - \theta_0\| = O_{\mathbb{P}}(1/\sqrt{n})$, so $\tilde{\theta}_n \xrightarrow{\mathbb{P}} \theta_0$. By continuity of $\nabla_\theta \Phi$ and the continuous mapping theorem, $\nabla_\theta \Phi(\tilde{\theta}_n) \xrightarrow{\mathbb{P}} \nabla_\theta \Phi(\theta_0)$. The result follows from Slutsky's lemma: the product of a term converging in distribution and a term converging in probability to a constant converges in distribution to their product.