The key is to differentiate the log twice and take the expectation. Applying the quotient rule, the Hessian of $\log f(X,\theta)$ with respect to $\theta$ is \begin{align*} \nabla^2_\theta \log f(X,\theta) = \nabla_\theta\!\left(\frac{\nabla_\theta f(X,\theta)}{f(X,\theta)}\right) = \frac{\nabla^2_\theta f(X,\theta)}{f(X,\theta)} - \frac{\nabla_\theta f(X,\theta)\, \nabla_\theta f(X,\theta)^\top}{f(X,\theta)^2}. \end{align*} Taking the expectation under $\mathbb{P}_\theta$ and using the regularity assumption to exchange differentiation and integration: \begin{align*} \mathbb{E}_\theta\!\left[\nabla^2_\theta \log f(X,\theta)\right] &= \int_{\mathcal{X}} \frac{\nabla^2_\theta f(x,\theta)}{f(x,\theta)}\, f(x,\theta)\, dx - \mathbb{E}_\theta\!\left[\frac{\nabla_\theta f(X,\theta)\, \nabla_\theta f(X,\theta)^\top}{f(X,\theta)^2}\right] \\ &= \nabla^2_\theta \int_{\mathcal{X}} f(x,\theta)\, dx - \mathbb{E}_\theta\!\left[\nabla_\theta \log f(X,\theta)\, \nabla_\theta \log f(X,\theta)^\top\right] \\ &= 0 - I(\theta). \end{align*} The first integral vanishes because $\int_{\mathcal{X}} f(x,\theta)\, dx = 1$, and differentiating a constant twice gives zero. Rearranging gives $I(\theta) = -\mathbb{E}_\theta[\nabla^2_\theta \log f(X,\theta)]$.