It is easily seen by substitution that if $\vec {B} = \text {curl} \vec {A}$ , then $\text {div} \vec {B}=0$. Can we prove the converse as well?
Let $\vec {B}=(B_x,B_y,B_z)$. Then,
$$ \frac {\partial B_x}{\partial x}+\frac {\partial B_y}{\partial y}+\frac {\partial B_z}{\partial z}=0 $$
This equation is valid for, $$ \frac {\partial B_x}{\partial x}=Z-Y, \frac {\partial B_y}{\partial y}=X-Z, \frac {\partial B_z}{\partial z}=Y-X,\ \ \ \ \ \ \ \ \ \ \ (1) $$
Where $X,Y,Z$ are arbitrary quantities whatsoever. Now, $$ X=\frac {\partial B_y}{\partial y}+Z=-\frac {\partial B_z}{\partial z}+Y $$ This implies that $X$ can be expressed as a $2^{\text {nd}}$ order partial derivative of some quantity, say $A_x$, since the denominator has both $\partial y$ and $\partial z$. Hence, $$ X= \frac {\partial ^2 A_x }{\partial y\ \partial z},Y= \frac {\partial ^2 A_y }{\partial z\ \partial x}, Z= \frac {\partial ^2 A_z }{\partial x\ \partial y} $$ Solving the set (1) of equations, we have, $$ B_x= \left( \frac {\partial A_z}{\partial y}-\frac {\partial A_y}{\partial z} \right),...\text{and 2 similar expressions.} $$
This is my line of thinking - I have made use of the 'form' of the differential-fractions. Yet I feel there are logical holes in the above outline. I need some suggestions, and references as well for a proper proof of the theorem. Thanks!
There are several questions about this already on this site. The answer is "it depends". It depends on the domain of $B$. If for example, the domain of $B$ is all of $\Bbb{R}^3$ or say an open ball, or more generally any star-shaped open set (I'm sure you can allow for more general situations as well) then yes the converse is true (look up Poincare's lemma for a proof).
BUT, we need some sort of restriction on the domain of the function because otherwise it is not true, as seen by the following counterexample: let $U=\Bbb{R}^3\setminus\{0\}$ and define $\mathbf{B}:U\to\Bbb{R}^3$ as $\mathbf{B}(\mathbf{r})=\frac{\mathbf{r}}{r^2}$. Then, $\mathbf{B}$ is divergence free, but doesn't admit a vector potential since the integral over the unit sphere is $4\pi\neq 0$ (if $\mathbf{B}$ did admit a vector potential on $U$ then by Stokes' theorem the integral must be $0$ since the unit sphere has no boundary, but it is not, hence there is no vector potential on $U$).