Having a potential function and being conservative are equivalent (under some mild assumptions).
Specifically, if a (continuous) vector field is conservative on an open connected region then it has a potential function.
And "Yes" if a vector field fails to be conservative, it cannot have a potential function.
Here are some notes I posted for one of my classes a few years ago...
http://mathsci2.appstate.edu/~cookwj/courses/math2130-fall2009/math2130-Line_Int_notes.pdf
A few notes:
1) I didn't list all assumptions everywhere (for example, I wasn't careful to say that I'm assuming things are continuous where needed).
2) In the notes a vector field which possesses a potential function is called a "gradient" vector field.
3) The relevant theorem is on page 5.
You don't have to find the integration constant immediately. Keep proceeding as follows.
After you determined that $f(x,y,z) = xy+g(y,z)$, differentiate with respect to $y$.
This gives $\frac{\partial f}{\partial y}=x+\frac{\partial g}{\partial y}=F_y=x$.
Thus, $\frac{\partial g}{\partial y}=0$, which implies that $g$ is a function of $z$ only. In turn, this means that $f(x,y,z)=xy+h(z)$.
Next, differentiate $f$ with respect to $z$.
This gives $\frac{\partial f}{\partial z}=h'(z)=F_z=z^2$.
Thus, $h(z)=\frac13z^3+C$.
Finally, $f(x,y,z)=xy+h(z)=xy+\frac13z^3+C$.
To check this, we have
$$\vec F=\nabla f(x,y,z)$$
$$=\hat x\frac{\partial f}{\partial x}+\hat y\frac{\partial f}{\partial y}+\hat z\frac{\partial f}{\partial z}$$
$$=\hat xy+\hat yx+\hat zz^2$$which completes the task!
Best Answer
If we manage to find a potential function $U$ for $\mathbf{F}$, we are done so let's try to do that. We need:
$$ \frac{\partial U}{\partial x} = 2x + y \implies U(x,y,z) = \int (2x + y) \, dx + G(y,z) = x^2 + yx + G(y,z) $$
for some function $G = G(y,z)$ which depends only on $y,z$. Next, $$ \frac{\partial U}{\partial y} = z \cos(yz) + x \implies x + \frac{\partial G}{\partial y} = x + z \cos(yz) \implies \\ G(y,z) = \int z \cos(y z) \, dy + H(z) = \sin(yz) + H(z)$$
for some function $H = H(z)$ that depends only on $z$.
Finally,
$$ \frac{\partial U}{\partial z} = y \cos(yz) \implies y \cos(yz) + \frac{\partial H}{\partial z} = y \cos(yz) \implies H \equiv C $$
so a potential for $\mathbf{F}$ is given by
$$ U(x,y,z) = x^2 + yx + \sin(yz) + C $$
where $C \in \mathbb{R}$ is arbitrary.