= Solution
Write $\ell(x,\theta)=\log f(x,\theta)$ and let $\dot\ell_{\theta_0}(x)$ be its <score function>. A second-order <Taylor expansion> around $\theta_0$ gives, for fixed $h$,
$$
Z_n(h)=h^T\Delta_n+\frac1{2n}\sum_{i=1}^nh^T\ddot\ell(X_i,\theta_{i,n})h,
\qquad
\Delta_n=\frac1{\sqrt n}\sum_{i=1}^n\dot\ell_{\theta_0}(X_i),
$$
where every $\theta_{i,n}$ lies between $\theta_0$ and $\theta_0+h/\sqrt n$. The <mean-zero score identity>, the <Fisher information matrix>, and the <multivariate central limit theorem> give
$$
\Delta_n\xrightarrow{d}N(0,I(\theta_0)).
$$
The assumed regularity and a <uniform law of large numbers> give
$$
\frac1n\sum_{i=1}^n\ddot\ell(X_i,\theta_{i,n})\xrightarrow{p}-I(\theta_0).
$$
Consequently, by the <Slutsky theorem>,
$$
Z_n(h)\xrightarrow{d}h^T\Delta-\frac12h^TI(\theta_0)h,
\qquad
\Delta\sim N(0,I(\theta_0)).
$$
For the stated <Gaussian shift model>, direct expansion of the two <normal distribution> densities gives
$$
Z(h)=h^TI(\theta_0)X-\frac12h^TI(\theta_0)h,
\qquad
X\sim N(0,I(\theta_0)^{-1}).
$$
Since $I(\theta_0)X\sim N(0,I(\theta_0))$, this has exactly the limiting distribution above. Equivalently, both limits are normal with mean $-\tfrac12h^TI(\theta_0)h$ and variance $h^TI(\theta_0)h$. This is the <local asymptotic normality> of the model at $\theta_0$.
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