Conditions for a vector field to be a gradient Consider a vector field $F : \mathbf{R}^d \to \mathbf{R}^d$. 
I want to know standard conditions on $F$ which guarantee that $F$ is a gradient, i.e. that $F = \nabla V$ for some function $V$. 
I know that gradient fields (sufficiently smooth, etc. etc.) all satisfy
$$\frac{ \partial F_i }{ \partial x_j} = \frac{ \partial F_j }{ \partial x_i} \quad \text{for all } i, j$$
and I have it in my mind that this is also a sufficient condition (or pretty close to it), but I don't really trust myself on this, and have no reference to this effect. 
Is this the case? Is there another sufficient condition? What is a good reference for this sort of thing?
 A: If you're considering $C^1$ vector fields $F$ defined on all of $\Bbb{R}^n$, then being a gradient field is equivalent to 
\begin{align}
\dfrac{\partial F_i}{\partial x_j} &= \dfrac{\partial F_j}{\partial x_i} \quad \text{for all $i,j \in \{1, \dots, n\}$}\tag{$*$}
\end{align}
As you've already mentioned, if $F = \nabla \phi$ is a gradient field (for some $C^2$ function $\phi: \Bbb{R}^n \to \Bbb{R}$) then the above equality is simply true by the equality of mixed partial derivatives.
To prove the converse statement, suppose $F: \Bbb{R}^n \to \Bbb{R}^n$ is $C^1$ and satisfies $(*)$. Now, define a function $\phi: \Bbb{R}^n \to \Bbb{R}$ by
\begin{align}
\phi(x) &:= \sum_{a=1}^n x_a \int_0^1 F_a(tx) \, dt
\end{align}
Then, a few lines of calculations (product rule, chain rule, differentiation under integral sign etc) shows that $\nabla \phi = F$. I suggest you first try to to compute it yourself; if you get stuck, look up the proof of Poincare's Lemma online, or let me know, I'll add some more details.
Anyway, note that the place where I used that $F$ is well-defined on all of $\Bbb{R}^n$ is when I used the integral $\int_0^1F_a(tx) \, dt$. In order for this integral to be well-defined, we need to ensure that the entire line-segment from $0$ to $x$ lies in the domain of $F$. So, actually, you can weaken hypotheses on the domain slightly: it is enough to assume that the domain, $U$, of $F$ is open in $\Bbb{R}^n$ and "star-shaped with respect to the origin", which means for every $x \in U$, the line segment from $0$ to $x$ lies in $U$ as well.
By the way, I'm not sure if "star-shaped with respect to origin" is the weakest possible assumption, but this is already a good enough start, so I'll leave it at that.

Also, if you know about differential forms, then you can rephrase your question as follows:

Let $\omega$ be a $C^1$, differential $1$-form defined on an open subset $U \subset \Bbb{R}^n$. Then, what are necessary and sufficient conditions on $\omega$ to ensure that $\omega$ is exact?

The answer is that it is necessary for $\omega$ to be closed ($d \omega = 0$). Poincare's lemma tell's you that a sufficient condition for $\omega$ to be exact is that $U$ be star-shaped (with respect to some particular point in $U$), and that $\omega$ be closed.
