First treat $n = 1$. Apply the Cauchy-Schwarz inequality to the random variables $Y = \tilde{\theta}(X)$ and $Z = \frac{d}{d\theta} \log f(X, \theta)$, which gives $\operatorname{Cov}(Y, Z)^2 \leq \operatorname{Var}(Y)\operatorname{Var}(Z)$, hence $\operatorname{Var}_\theta(\tilde{\theta}) \geq \operatorname{Cov}_\theta(\tilde{\theta}, Z)^2 / \operatorname{Var}_\theta(Z)$. Since $\mathbb{E}_\theta[Z] = 0$ (score has mean zero) and $\operatorname{Var}_\theta(Z) = I(\theta)$ by definition of Fisher information, the denominator is $I(\theta)$. For the numerator, compute \begin{align*} \mathbb{E}_\theta\!\left[\tilde{\theta} \cdot \frac{d}{d\theta} \log f(X, \theta)\right] = \int_{\mathcal{X}} \tilde{\theta}(x) \cdot \frac{\frac{d}{d\theta} f(x, \theta)}{f(x, \theta)} \cdot f(x, \theta)\, dx = \frac{d}{d\theta} \int_{\mathcal{X}} \tilde{\theta}(x) f(x, \theta)\, dx = \frac{d}{d\theta} \theta = 1, \end{align*} where the interchange of integral and derivative uses the regularity assumption, and the final step uses unbiasedness $\mathbb{E}_\theta[\tilde{\theta}] = \theta$. Thus $\operatorname{Cov}_\theta(\tilde{\theta}, Z) = 1$, giving $\operatorname{Var}_\theta(\tilde{\theta}) \geq 1/I(\theta)$. For general $n$, replace $Z$ by $\frac{d}{d\theta} \log f(X_1, \ldots, X_n, \theta)$; its variance is $I_n(\theta) = nI(\theta)$ by the tensorization result, and the same calculation shows $\mathbb{E}_\theta[\tilde{\theta} Z] = 1$, yielding $\operatorname{Var}_\theta(\tilde{\theta}) \geq 1/(nI(\theta))$.