= Solution
Consider the subset
$$
A=\{x\in M:f_1(x)=f_2(x),\ (df_1)_x=(df_2)_x\}.
$$
It is nonempty by the given initial data. A <local isometry> preserves the <Levi-Civita connection> and therefore affinely parametrized <geodesics>. On a sufficiently small <normal neighborhood> of $x\in A$, this gives
$$
f_i(\exp_xv)=\exp_{f_i(x)}((df_i)_xv).
$$
The right sides agree, so $f_1=f_2$ on that neighborhood and their <differentials> agree there as well. Consequently $A$ is <open>.
It is also <closed>. If $x_j\in A$ converges to $x$, continuity first gives $f_1(x)=f_2(x)$. Choose coordinate charts around $x$ and this common image. The coordinate matrices of the two <differentials> depend continuously on the base point; their equalities at $x_j$ therefore pass to $x$. Thus $x\in A$. Since $M$ is <connected>, the nonempty subset $A$ that is both open and closed equals $M$. Hence \b[$f_1=f_2$ everywhere]. This proves <local isometries are determined by first-order data> without requiring completeness.
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