Let $ X $ and $ Y $ be topological spaces and let $ f\colon X\to Y $ be continuous. Let's suppose that all I know about the inverse image functor $$ f^*\colon \mathrm{Sh}(Y)\to \mathrm{Sh}(X) $$ (where $ \mathrm{Sh}(X) $ is the category of sheaves on $ X $) is that $ f^* $ is the left adjoint of the obvious direct image sheaf functor $ f^*\colon \mathrm{Sh}(X)\to \mathrm{Sh}(Y) $.
Is it possible to show that for every sheaf $ \mathscr G $ on $ Y $ and for every $ x\in X $ it is exactly $ (f^*\mathscr G)_x \cong \mathscr G_{f(x)} $ without appealing to any explicit construction of $ f^*\mathscr G $?
P.S. Of curse I know that defining $ f^* $ to be the composite of $$ \mathrm{Sh}(Y)\xrightarrow{\Lambda}\mathrm{\acute{E}t}(Y)\xrightarrow{\text{pullback along $ f $}} \mathrm{\acute{E}t}(X)\xrightarrow{\Gamma}\mathrm{Sh}(X) $$ then $ f^*\dashv f_* $, where $ \Lambda $ and $ \Gamma $ are the well-known étale space and sheaf-of-section functors. But I was looking for a more conceptual way to prove this fact
We can use the fact that the stalk functor ${-}_x$ has a right adjoint which is the skyscraper sheaf functor $i_{x*}$, where $$(i_{x*}G)(U) := \begin{cases} G, & x \in U; \\ 0, & x \notin U \end{cases}$$ and the restriction map is either the identity map if $x \in V \subseteq U$, or the zero map otherwise.
Therefore, also using the mentioned adjunction between $f^*$ and $f_*$, we get that ${-}_{x} \circ f^*$ is left adjoint to $f_* \circ i_{x*}$. However, it is easy to show that we have an isomorphism of functors $f_* \circ i_{x*} \simeq i_{f(x)*}$, which has left adjoint ${-}_{f(x)}$. Hence, the uniqueness of left adjoints gives a canonical isomorphism of functors ${-}_x \circ f^* \simeq {-}_{f(x)}$.