Metric of a cross-section in General Relativity Consider a finite closed region $V=(x,y,z)$ as a simply-connected subset of a 3-dimensional flat Euclidean space ${\Bbb R}^3$ with the metric $\text{d}s^2=\text{d}x^2+\text{d}y^2+\text{d}z^2$.
A surface defined by a single-valued non-singular twice differentiable function $z=F(x,y)$ represents a cross-section of the region $V$, i.e. for any $(x,y)$ in this region, $z=F(x,y)$ is defined and contained in the region $V$.
The curvature of $F$ defined this way must have certain limitations, because in general, a 2-dimensional surface with an arbitrary intrinsic curvature cannot always be represented as a cross section of ${\Bbb R}^3$. (I am avoiding the term "embedding" to keep things simple.)
In other words, the metric (or curvature) of $F$ must obey certain conditions to allow $F$ to be a cross-section of $V$. Can these conditions be mathematically formulated? What limitations must be put on the metric (or on the curvature) of a 2-space to allow defining this space as a cross-section of ${\Bbb R}^3$?

 A: In 1971 Jacobowitz proved that it is always possible to locally isometrically embed a surface in $\mathbb R^4$. 
As mentioned in its introduction, it is not exactly known when a local isometric embedding into $\mathbb R^3$ is possible, though it is around points where the curvature is nonzero (see e.g. the book by Han and Hong, chapter 4), and whenever the metric is analytic (this was proved by Janet in 1926, see Wikipedia). 
Subsequent chapters of the same book show cases in which it is know to hold around points of curvature 0 for a non-analytic metric: in chapter 5 that it holds when the metric is of class $\mathcal C^9$ if the gradient of the curvature is nonzero. Chapter 6: when $g$ is $\mathcal C^{12}$ and the curvature is nonnegative around the point. Chapter 7 treats the non-positive case under a more technical condition than just the differentiability class.
It is an open problem if every smooth surface can be locally isometrically embedded in $\mathbb R^3$ with the Euclidean metric, but as the preceding results show, it will be very hard to construct a counterexample.
EDIT
Since some clarifications in the comments were moved to chat, I'll briefly recap the question as I interpreted it, based on them. 
We consider surfaces that are smooth submanifolds of $\mathbb R^3$ with the induced metric, and the question is which abstract surfaces can locally occur as such surfaces.
The answer essentially is: every surface you can think of, and maybe every surface.
A: Answer based on discussion here. 
Let's represent the Schwarzschild metric in $(n+1)$ dimensions with a $(-+++)$ metric signature,
\begin{equation}
     ds^{2}=-\left(1-\frac{r_s}{r^{n-2}} \right)dt^{2} +\left(1-\frac{r_s}{r^{n-2}} \right)^{-1} dr^{2} +r^{2} d\Omega^{2}.
\end{equation}
Note that here $d\Omega^2$ is the round unit metric on a sphere $S^{n-1}$. Following the embedding machinery, we shall embed this metric (it's time slice) in a $(n+1)$-dimensional Euclidean space as follows
\begin{equation}
\begin{array}{lr}
    ds^2=dz^{2}+(dx^{1})^{2}+(dx^{2})^{2}+...+(dx^{n})^{2}=\left[\left(\frac{dz}{dr}+1\right)\right]^{2}dr^{2}+r^{2} d\Omega^{2}.
    \end{array}
\end{equation}
This equation will coincide with the spacelike component of the time-sliced $(n+1)$-dimensional Schwarzschild metric. Thus, a smooth function is obtained.
\begin{equation}
    \frac{dz}{dr}=\pm \sqrt{\frac{r_s}{r^{n-2}-r_s}}.
\end{equation}
Note that this function can be explicitly integrated in $3$ and $4$ dimensions. It was Kasner who showed that it was impossible to embedd a four-dimensional manifold in a five-dimensional flat space which can be found here. Another way of interpreting this is to observe that the geometry described by the function of the surface of revolution doesn't replicate the Flamm paraboloid. Things get a bit complicated in $n\ge 5$ dimensions as $z(r)$ is expressed in terms of elliptic functions. The qualitative behaviour in this dimension, or for that matter in $n\ge 5$ dimensions is quite different, here $z(r)$ asymptotically diminishes to a finite value as $r\rightarrow \infty$. The smooth function takes the following form in $(n=5)$ dimensions
\begin{equation}
    \frac{dz}{dr}=\pm \sqrt{\frac{r_s}{r^3-r_s}}.
\end{equation}
Integration yields the following result (to simplify calculations let us make the assumption that $m=1$)
\begin{equation}
    z(r)= \pm  \frac{2i}{3^{\frac{1}{4}}}\sqrt{(-1)^{\frac{5}{6}}(r-1)\frac{r^{2}+r+1}{r^{3}-1}}F\left(sin^{-1}\left(\frac{\sqrt{-ir-(-1)^{\frac{5}{6}}}}{3^{\frac{5}{6}}} \right) \right)(-1)^{\frac{1}{3}},
\end{equation}
where $F(x|m)$ is the elliptic integral of the first kind with parameter $m=k^{2}$.
Wave coordinates are another set of coordinates that act as an effective tool for PDE analysis of spacetimes. In spherical coordinates associated to wave coordinates $(t, \hat{x}, \hat{y}, \hat{z})$, with radius function $\hat{r} = \sqrt{\hat{x}^{2} + \hat{y}^{2} + \hat{z}^{2}}$, the Schwarzschild metric takes the following form (this is obtained by replacing $r$ in the Schwarzschild metric with $\hat{r} = r - m$)
\begin{equation}
    ds^{2} = - \frac{\hat{r} - m}{\hat{r} + m}dt^{2} + \frac{\hat{r} + m}{\hat{r} - m}d\hat{r}^{2} + \left(\hat{r} + m \right)^{2}d\Omega^{2}.
\end{equation}
Now, consider the Schwarzschild metric in dimensions greater than $3$, i.e., $n\geq 3$,
\begin{equation}
    ds^{2}_{n} = -\left(1 - \frac{2m}{r^{n-2}} \right)dt^{2} + \left(1 - \frac{2m}{r^{n-2}} \right)^{-1} dr^{2} + r^{2}d\Omega^{2},  
\end{equation}
and consider a general spherically symmetric static metric of the the following form
\begin{equation}
    ds^{2} = -e^{2\alpha}dt^{2} + e^{2\beta}d\hat{r}^{2} + e^{2\gamma}\hat{r}^{2}d\Omega^{2},
\end{equation}
where $\alpha$, $\beta$, and $\gamma$ depend only upon $r$. Now let us define $\phi$ and $\psi$ such that
\begin{equation}
    \phi = e^{\alpha + \beta + (n-3)\gamma}, \ \ \ \psi = e^{\alpha + \beta + (n-1)\gamma}\left(e^{-2\beta} - e^{-2\gamma}  \right).
\end{equation}
Now, we proceed to perform all our calculations in a coordinate system in which the vector $(x,y,z)$ is aligned along the $x$-axis, $(x,y,z) = (r,0,0)$. Then the metric in a spacetime dimension of $n+1$ reads
\begin{equation}
    g = \left( \begin{array}{ccccc}
-e^{2\alpha} & 0 & 0 & 0 \cdots  & 0 \\ 
0  & e^{2\beta} & 0 & 0 \cdots  & 0 \\ 
0  & 0 & e^{2\gamma} & 0 \cdots  & 0 \\ 
0  & 0 & \vdots &  \ddots  & 0  \\ 
0  & 0  & 0 & \cdots & e^{2\gamma} \end{array}\right),
\end{equation}
which implies the following
\begin{equation}
    det \ g = |g| = e^{2(\alpha + \beta) + 2(n-1)\gamma},
\end{equation}
still at $(x,y,z) = (r,0,0)$. Spherical symmetry implies that this equality holds everywhere. Now, we have
\begin{equation}
\begin{array}
    \sqrt{det\ g}\  \partial_{g}\partial^{g}x = \partial_{\xi}\left(\sqrt{|g|} g^{\nu \mu} \right)\\\\    
 = \partial_{\xi} \left(e^{-2\gamma}\delta^{\mu \nu} + \left(e^{-2\beta} - e^{-2\gamma} \right)\frac{x^{\mu}x^{\nu}}{r^{2}} \right)\\\\
 = \partial_{\xi} \left(\underbrace{e^{\alpha + \beta + (n-3)\gamma}}_{= \phi}\delta^{\mu \nu} + \underbrace{e^{\alpha + \beta + (n-1)\gamma}\left(e^{-2\beta} - e^{-2\gamma}  \right)}_{ = \psi} \frac{x^{\mu}x^{\nu}}{r^{2}} \right)\\\\
 = \left(\phi^{'} + \psi^{'} + \frac{n-1}{r} \psi \right)\frac{x^{\mu}}{r},
\end{array}
\end{equation}
this is called the harmonicity condition. It follows from this condition that 
\begin{equation}
  0 = \frac{d\left(\phi + \psi \right)}{d\hat{r}} + \frac{n-1}{\hat{r}}\psi,   
\end{equation}
equivalently,
\begin{equation}
    \frac{d\left[\hat{r}^{n-1}\left(\phi + \psi \right) \right]}{d \hat{r}} = \left(n-1 \right)\hat{r}^{n-2}\phi.
\end{equation}
Now, comparing the above equation with the standard metric defined, we find
\begin{equation}
    e^{\alpha} = \sqrt{1 - \frac{2m}{r^{n-2}}}, \ \ e^{\beta} = \frac{dr}{d\hat{r}} e^{-\alpha}, \ \ e^{\gamma} = \frac{r}{\hat{r}}.
\end{equation}
Since $\phi + \psi = e^{\alpha - \beta + (n-1)\gamma}$ and $\phi = \frac{dr}{d\hat{r}}\left(\frac{r}{\hat{r}} \right)^{n-3}$, we finally obtain (which upon introduction of $x = 1/r$ becomes)
\begin{equation}
    \frac{d}{dr}\left[x^{3-n} \left(1 - 2mx^{n-2} \right) \frac{d\hat{r}}{dx} \right] = (n-1)\hat{r}x^{1-n}
\end{equation}, this is called an equation with a Fuchsian singularity at x = 0
\begin{equation}
 \frac{d}{dr}\left[r^{n-1} \left(1 - \frac{2m}{r^{n-2}} \right) \frac{dr}{d\hat{r}} \right] = (n-1)\hat{r}r^{n-3}.
\end{equation}
The characteristic exponents are $−1$ and $n-1 $ so that, after matching a few leading coefficients, the standard theory of such equations provides solutions with the behaviour
\begin{equation}
\hat{r} = r - \frac{m}{\left(n-2 \right)r^{n-3}} + \left\{\frac{m^{2}}{4}r^{-2}\ ln(r) + \mathcal{O}\left(r^{-5}\ ln(r) \right), n = 4\\
\mathcal{O}\left(r^{5 - 2n} \right)  , n \geq 5.\right\}
\end{equation}
Thus, somewhat surprisingly, we find logarithms of $r$ in an asymptotic expansion of $\hat{r}$ in dimension $n = 4$. However, for $n\geq 5$ there is a complete expansion of $\hat{r}$ in terms of inverse powers of $r$, without any logarithmic terms in those dimensions.
