= Solution
A subset $X\subset M^m$ is a $d$-dimensional <embedded submanifold> if, with the subspace topology, every point has a smooth ambient chart in which $X$ is the intersection of the coordinate domain with $\mathbb R^d\times\{0\}$. These slice charts give its smooth structure, and its inclusion is an <embedding>.
The <inverse function theorem> says that a smooth map $F:U\subset\mathbb R^m\to\mathbb R^m$ with invertible derivative at $a$ restricts to a <diffeomorphism> between open neighborhoods of $a$ and $F(a)$. A point $q\in N$ is a <regular value> of a smooth map $f:M^m\to N^n$ when $df_x:T_xM\to T_qN$ is surjective at every $x\in f^{-1}(q)$.
Choose local coordinates centered at $x$ and $q$. At a fiber point, rank $n$ means that, after relabeling coordinates, the first $n$ columns of the coordinate derivative form an invertible <matrix>. Define
$$
F(u_1,\ldots,u_m)=\bigl(f_1(u),\ldots,f_n(u),u_{n+1},\ldots,u_m\bigr).
$$
Its derivative is block upper triangular with that invertible block and an identity block, hence invertible. The inverse function theorem makes $F$ a new ambient coordinate chart in which $f^{-1}(q)$ is exactly $v_1=\cdots=v_n=0$. Therefore the <regular level set theorem> gives
$$
\boxed{f^{-1}(q)\text{ is an embedded submanifold of dimension }m-n}
$$
when the fiber is nonempty. An empty fiber is regular vacuously; if $m<n$, it is the only possible regular fiber. The construction does not assign a negative dimension to a nonempty manifold.
For the projective hyperplane, the map
$$
\iota:\mathbb{RP}^1\longrightarrow\mathbb{RP}^2,\qquad[a:b]\longmapsto[a:b:0]
$$
is injective with inverse $[x_0:x_1:0]\mapsto[x_0:x_1]$ on its image. At every image point, either $x_0$ or $x_1$ is nonzero; the corresponding standard projective chart uses $x_2/x_i$ as one coordinate and cuts the image out by setting it to zero. The map and inverse are smooth in these charts, and the slice condition proves
$$
\boxed{X\text{ is embedded and }X\cong\mathbb{RP}^1}.
$$
Nevertheless, $\mathbb{RP}^2\setminus X$ is the single affine chart $x_2\ne0$, diffeomorphic to $\mathbb R^2$, so it is connected. If a smooth real function had exactly this zero set with zero regular, the inverse function theorem near any point of $X$ would give both positive and negative values off $X$. On the connected complement it never vanishes, so continuity requires its sign to be constant. This contradiction is the obstruction that <regular zero hypersurfaces have disconnected complements>. Thus
$$
\boxed{X\text{ cannot be a global regular zero set of a smooth real function}}.
$$
It illustrates why local defining equations for an embedded hypersurface need not glue to a global real defining function, as in <real projective hyperplane is not a global regular zero set>.
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