Let $M$ be a manifold of dimension $2n$, $f:M\rightarrow\mathbb{R}^n$ a smooth map.
We know that if $f$ is a submersion, then for any $c\in f(M)$, $f^{-1}(c)$ is a $n$-submanifold of $M$.
But what about the converse, if $f$ is surjective and if every level set of $f$ is a $n$-submanifold of $M$, does that imply that $f$ is a submersion?
No, the converse is false. To come up with a counterexample, let us take $M$ to be a vector space $\Bbb{R}^m$. Note that if $f$ were linear, then the converse is true (rank-nullity), so to get a counterexample, we need to use non-linear functions.
So, consider $f:\Bbb{R}^2\to\Bbb{R}$ defined as $f(x,y)=x^3-y^3$, which is clearly surjective, and has $f'(x,y)= (3x^2, -3y^2)$ which vanishes only at the origin; so $f$ is surjective but not a submersion. Now, for any $c\in\Bbb{R}\setminus\{0\}$, we have that $(0,0)\notin f^{-1}(\{c\})$ and thus at each $p\in f^{-1}(\{c\})$, we have $f'(p)$ being non-zero and thus $f^{-1}(\{c\})$ is an embedded 1-dimensional submanifold of $\Bbb{R}^2$. Also, the level set $f^{-1}(\{0\})$ is just the line $\{(x,y)\in\Bbb{R}^2\,:\, x=y\}$ which is clearly a 1-dimensional embedded submanifold. Hence, $f$ provides a counterexample to the converse.