How to determine if Vector Function is a possible Electric Field? So there are plenty of examples online (including e.g. this one) on how to determine if there is a possible electric field for vectors where $z=0$ 
However I am given 
\begin{align}
\vec E & = xy  \hat i + 2yz \hat j + 3xz \hat k \quad\text{and}\\
\vec E& =y^2 \hat i+(2xy+z^2) \hat j+2yz \hat k
\end{align}
Where $\hat i$, $\hat j$ and $\hat k$ are the unit vectors to show what direction. 
Do I a triple integral or do I even just ignore z? I can't find any resources on what to do when $z$ is not equal to $0$
 A: A vector field is a possible electric field in the electrostatic regime if and only if its curl is zero: i.e. if and only if
$$
\nabla \times\mathbf E = 
\begin{pmatrix}
\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} \\
\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \\
\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y}
\end{pmatrix}
=\mathbf 0.
$$
If you calculate all the (differences of) partial derivatives and they give zero, then it is a valid electric field.
For more details, see your favourite introductory electromagnetism textbook.

The above is an exact criterion and there is absolutely no reason to demand another one. However, if you really, absolutely, because of personal preference, must have an integral-based criterion, then a necessary and sufficient condition is that the line integral
$$
\int_{\mathbf 0}^{\mathbf r} \mathbf E\cdot\mathrm d\mathbf l,
$$
where you integrate along line segments parallel to the coordinate axes, be independent of which permutation of such segments you take, over all such possible permutations (six in total). This is much more work, though - seriously, unless your problem sheet explicitly demands it or something, use the differential criterion. 
A: If you are in a electrostatic situation, the electric field ought to be conservative, as you seem to imply in your suggestion of the triple integral.
A faster way to check if a field is conservative is to calculate its rotational.
Any sufficiently regular field$^1$ whose rotational is zero is also a conservative field.
Since all your fields have infinitely many continuous derivatives, this result aplies, and we can simply calculate:$$
\nabla \times E=\begin{vmatrix}
i & j & k\\ 
\frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z}\\ 
E_x & E_y & E_z \\
\end{vmatrix}$$
And see wether or not it is zero.
None of your examples yield zero, and thus, they are both not conservative, and unfit to represent electrostatic fields.
However, is the situation is not electrostatic, no restrictions that i am aware of exists, and any vector field could represent an electric field.
$^1$ In this context, any continuous function with continuous partial derivatives is regular enough
