Q:

How to solve vector equations?

I don't know what is meant when asked the following;
I am asked, for the vector function F determine wether the equation $\nabla \phi = F$ has a solution and determine $\phi$ .
for
$(2xyz^3 ,-x^2z^3-2y, 3x^2yz^2)$
But I don't know what is being asked.
Am I supposed to calculate the partial derivatives with respect to x, y and z for each component?
For example, should I find 
$\partial/\partial x = 2yz^{3}$
$\partial/\partial y= -2$
$\partial/\partial z= 6x^{2}yz$
Or do I integrate each component or something? I am just confused on what is going on. Please help to explain.
Of course I want to understand this specific example but I mostly want to understand in general what is going on. Thank you
Update: From the comments I am seeing I am really just lost. I think I need to see it actually worked out. It is only one of many more examples I will do later but I am lost

A:

You are being asked to find a function $\phi$ such that the vector field $F$ that you have been given is the gradient of $\phi$. We know that
$$\nabla \phi = \frac{\partial \phi}{\partial x} \hat{i} + \frac{\partial \phi}{\partial y} \hat{j} + \frac{\partial \phi}{\partial z} \hat{k}$$
and 
$$ F = (2xyz^{3}) \hat{i} - (x^2z^3 + 2y)\hat{j} + (3x^2yz^2)\hat{k} $$
If we set $\nabla \phi = F$, we get three differential equations:
$$ \frac{\partial \phi}{\partial x} = 2xyz^{3}$$
$$ \frac{\partial \phi}{\partial y} = -x^2z^3 - 2y$$
$$ \frac{\partial \phi}{\partial z} = 3x^2yz^2$$
These can all be solved using partial integration. You can then combine the results to find a function $\phi$ such that $\nabla \phi = F$ (if such a function exists). As user runaround mentioned, you can check existence of $\phi$ by checking if $\nabla \times F = 0$. If $\nabla \times F \neq 0$, then no such $\phi$ exists since a gradient vector field is always curl-free.
Can you take it from here?

