Let $f: \mathbb{R}^d \rightarrow \mathbb{R}$ be an analytic function, with infinite radius of convergence at any point $x \in \mathbb{R}^d$, where $d>1$ (and, in particular, $d=2$). It is clear that, if we are given the restriction of $f$ to any open set, this is enough to determine $f$ on the whole of $\mathbb{R}^d$, since we can just evaluate the Taylor series of $f$ at some $x_0$ in this open set.
Suppose we are now given the level lines of $f$ instead; $f^{-1}(\{0\}) := \{ x \in \mathbb{R}^d : f(x) = 0 \}$. Is this enough to determine $f$ uniquely (up to multiplicative constant)? What if we are only given one connected component of $f^{-1}(\{0\})$? Or even just the (nontrivial) intersection of one connected component of $f^{-1}(\{0\})$ with an open set? Are these enough to determine $f$ uniquely, modulo constant?
I have been told that all of these are enough to uniquely determine $f$, but have been unable to find a convincing reference or intution for why this might be the case, unlike when we view $f$ on an open set.