Yes, you want the graph in $\mathbb{R}^3$ of a function $f:\mathbb{R}^2\to\mathbb{R}$. The easy way to see that this is an embedded surface is to define the map $F:\mathbb{R}^2\to\mathbb{R}^3$ via $F(x,y) = (x,y,f(x,y))$ and check that $F$ is an embedding: it is
- injective,
- smooth, and
- $dF$ is everywhere full rank.
The first two are easy to check. The third follows from the tact that $dF = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \end{pmatrix}$ always has trivial kernel.
(In fact, the graph of a smooth function is actually a two-dimensional submanifold of $\mathbb{R}^3$, which is slightly stronger than noting that it is an embedded surface. Observe that we can produce new coordinates on $\mathbb{R}^3$ by "flattening" the graph down to the $xy$-plane. The change-of-coordinate map is $\phi(x,y,z) = (x,y,z - f(x,y))$; it's a diffeomorphism, which means that the graph of $f$ has a coordinate chart in which it is the $xy$-plane, i.e., it is a two-dimensional submanifold of $\mathbb{R}^3$.)