Let $f \in C^\infty_0(\mathbb{R}^3)$ be a smooth scalar function of compact support. I would like to find a smooth vector field $X : \mathbb{R}^3 \to \mathbb{R}^3$ satisfying $\operatorname{div}(X) = f$ such that $X$ is also of compact support. Is that possible?
Looking for $X$ of the form $X = \operatorname{grad} \Phi$ does not work, since solving the Laplace equation $\Delta \Phi = f$, $\Phi$ will not have compact support.
Here is a possible construction (provided the necessary condition that $f$ has zero integral is satisfied): Let $Y$ be such that $Y$ is a gradient and $div(Y)=f$. Now pick a ball B containing the support of $f$. On B, solve the Stokes problem $\Delta Z-\nabla p=0$, $div(Z)=0$, with the boundary condition $Z=-Y$. Now let $X=Y+Z$ inside B and $X=0$ outside B.
This is impossible in general. Indeed, if $X$ has a compact support, then the flux of $X$ through a large enough sphere is $0$, whereas the integral of $f$ over the corresponding ball (say $B$) will not be $0$ if e.g. the restriction of $f$ to $B$ is a nonnegative nonzero function, so that the integral of $f$ over $B$ be nonzero. So, if $\text{div}\,X=f$, we get a contradiction with the divergence theorem.
This has been answered well but as a long comment I would like to say that there is a field $X$ that is bounded by a constant multiple of $|x|^{-2}$. It is part of the Poincare Lemma (see Spivak's Calculus on Manifolds and convert from forms to vector notation; there is an exposition of that on my web page bterrell.net) that $$ f(x) = {\rm div\,}\Big(\int_0^1 t^2xf(tx)\,dt\Big). $$ In the case where $f$ has support in a ball of radius $b$, the integral only needs to run from $0$ to $\frac{b}{|x|}$, and is seen to be bounded by $$ \tfrac{1}{2}b |f|_{\rm max}\Big( \frac{b}{|x|} \Big)^2. $$
