If $X$ is metrisable and $f$ is a perfect surjection onto $Y$, where perfect means continuous, closed and pre-images of singletons are compact, then $Y$ is metrisable. This is proved e.g. in Engelking's General Topology (Thm. 4.4.15), and due independently to K. Morita and S. Hanai ("Closed mappings and metric space", Proc. Japan Acad. 32 (1956), 10-14) and A.H. Stone ("Metrizability of decomposition spaces", Proc. Amer. Math. Soc 7 (1956), 690-700). The proof in Engelking is based on the Bing-Nagata-Smirnov metrisation theorem.

A proper (in the sense of inverse images of compact sets being compact) continuous map with a $k$-space as codomain is closed, if the codomain is Hausdorff (for a simple proof see this paper) so the above theorem applies, as all locally compact spaces are $k$-spaces (or compactly generated); for a proof for locally compact see this question. This means that no counterexample exists.

As an afterthought, I looked a bit more at the question whether a proper continuous image of a metric space is metrisable, without conditions. We saw above that with $Y$ being a Hausdorff $k$-space (which are necessary conditions to be metrisable anyway) the answer is yes, as we then have a perfect map. But without a condition on $Y$ we do have an example: let $X$ be the reals in the discrete topology, and $Y$ the set of reals with the topology where all subsets of $\mathbb{R} \setminus \{0\}$ are open, and a set $O$ containing $0$ is open iff $O$ is co-countable (so it's essentially the one-point Lindelöfication of the reals in the discrete topology). One checks that $Y$ is hereditarily normal (but not perfectly normal) and is not a $k$-space because the only compact subsets of $Y$ are the finite ones (and so all subsets are compactly closed, but some are not closed). The last fact also implies that $f(x) = x$ is continuous and proper, but its image is not metrisable. And $X$ is locally compact metrisable.

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