Write the Taylor expansion $g(\hat{\theta}_n) = g(\theta_0) + g'(\theta_0)(\hat{\theta}_n - \theta_0) + o(|\hat{\theta}_n - \theta_0|)$ as $\hat{\theta}_n \to \theta_0$. Multiplying through by $\sqrt{n}$ gives $\sqrt{n}(g(\hat{\theta}_n) - g(\theta_0)) = g'(\theta_0)\cdot\sqrt{n}(\hat{\theta}_n - \theta_0) + o_{\mathbb{P}}(1)$. The remainder is $o_{\mathbb{P}}(1)$ because $\sqrt{n}(\hat{\theta}_n - \theta_0) = O_{\mathbb{P}}(1)$ and the $o(|\hat{\theta}_n - \theta_0|)$ term divided by $|\hat{\theta}_n - \theta_0|$ tends to zero. Slutsky's lemma applied to the product $g'(\theta_0) \cdot \sqrt{n}(\hat{\theta}_n - \theta_0)$ with the constant $g'(\theta_0)$ gives the result.