Is every smooth manifold the solution set of finitely many equations? In the case of affine varieties we have finitely many (polynomial) equations defining the variety. It is a (smooth) manifold iff it satisfies the Jacobian criterion.
I wonder whether this can be generalized in the following sense:
Is it possible for every smooth submanifold $M\subseteq \mathbb{R}^n$ to find a smooth map $f: \mathbb{R}^k \rightarrow \mathbb{R}^l$ and a regular value $q\in \mathbb{R}^l$ of $f$ such that $M=f^{-1}(q)$?
 A: First of all, your condition for smoothness for affine varieties is not quite right: It is sufficient but not necessary. You can say that a real  affine variety $X\subset R^n$  is smooth iff  for each point $x\in X$ there exists a Zariski open subset $X'\subset X$ containing $x$ such that 
$X'$ can be defined by a system of polynomial equations satisfying the Jacobian criterion. 
Now, suppose that $X\subset R^n$ is a submanifold which is the zero locus of a smooth mapping $f: R^n\to R^k$ such that $0$ is a regular value of $f$. Consider the normal bundle $N(X)$ in $R^n$. You can think of fibers $N_x$ of $N$ as orthogonal complements to $T_xX$. Then for each $x\in X$ the differential $df_x: N_x\to R^k$ is an isomorphism. This defines a trivialization of $N(X)$:
$$
(\pi, df): N(X)\to X\times R^k,
$$ 
where $\pi: N(X)\to X$ is the normal bundle. Therefore, the direct sum $T(X)\oplus N(X)$ is the trivial bundle over $X$ (the pull-back of the tangent bundle of $R^n$), which means that $T(X)$ is "stably trivial" and $X$ is "stably parallelizable". However, not every smooth manifold has stably trivial tangent bundle. For instance any nonorientable manifold $X$ gives an example of a manifold which cannot be realized as the zero locus of a smooth map to $R^k$ which has zero as the regular value. 
