It's well known that smooth embedded submanifold in $\Bbb{R}^n$ locally a level set (and locally is a graph), as in the thread Is every embedded submanifold globally a level set? setting up.
What if we consider only topological submanifold in $\Bbb{R}^n$? For the "locally graph" statement I see rectangles is a counterexample for the corner point cannot be locally graph. But for the "locally level set" statement I find no obivious counterexample.
Recall that given a topological spare $X$, a subset in $X$ is said to be precompact in $X$ if its closure in $X$ is compact.
And Lemma 1.10 in [LeeSM]: Every topological manifold has a countable basis of preompact coordinate balls.
Now given a topological manifold $M \subseteq \mathbb{R}^N$. From the above lemma, we have (countable many) open sets $U \subseteq \mathbb{R}^N$ covering $M$ such that $U \cap M$ is heomorphic to $\mathbb{R}^m$ and $\overline{U \cap M}$ is compact in $M$.
Then $\overline{U \cap M}$ is also compact in $\mathbb{R}^N$.
Hence $\overline{U \cap M}$ is closed (and bounded) in $\mathbb{R}^N$. By the strong Urysohn lemma, there is continuous function $$ f: \mathbb{R}^N \rightarrow[0,1] $$ such that $f(M)=0$ and $f>0$ otherwise.
Hence $U \cap M$ is a level set of $f$ in $U$.