$\newcommand{\F}{\mathbf{F}}\newcommand{\p}{\mathbf{p}}$If we consider the unbounded $\mathbb{R}^3$ case, there is a path integral formula to construct the right inverse of the curl operator for divergence free vector field
$$
\mathcal{R}(\F) = -(\p - \p_0)\times \int^1_0 \F\Big(\p_0 + t(\p- \p_0)\Big)t\,dt.\tag{1}
$$
A proof of this formula can be found here, for more discussion the author pointed to Spivak's book Calculus on Manifolds.
In your case, letting $\p_0 = \langle 0,0,0\rangle $, it is
$$
\mathcal{R}(\F) = \langle x,y,z\rangle \times \int^1_0 \langle y,z,x\rangle t^2 dt
= -\frac{1}{3}\langle z^2-xy,x^2-yz,y^2-xz\rangle .$$
You can check that this is indeed the right inverse
$$
\nabla \times \mathcal{R}(\F) = \F = -\langle y, z, x\rangle.
$$
Notice the vector field above is difference from yours, because essentially $\mathcal{R}(\F) + \nabla \phi$ is also the answer for smooth $\phi$.
Further checking: the difference between the right inverse $\mathcal{R}(\F) $ above and your potential field $\langle xy,0,−y^2/2+xz\rangle$ is
$$
\mathbf{A} = \langle \frac{2}{3}x y +\frac{1}{3}z^2, \frac{1}{3}(x^2-y z), -\frac{1}{6}y^2+\frac{2}{3}xz\rangle = \nabla \left(\frac{1}{3} x^2y + \frac{1}{3}xz^2 - \frac{1}{6}y^2 z\right),
$$
indeed a gradient.
If you wanna eliminate the huge kernel of curl operator, what you can do is (1) choosing a gauge (pinning down the divergence of $\mathbf{G}$), and/or (2) specifying boundary condition restricted on a simply-connected domain $\Omega$ of $\mathbb{R}^3$, by posing the following boundary value problem on some $\Omega$:
$$\left\{
\begin{aligned}
\nabla \times \mathbf{G} &= \F\quad \text{ in }\Omega,
\\
\nabla \cdot \mathbf{G} &= g \quad \text{ in }\Omega,
\\
\mathbf{G} \cdot \mathbf{n} &= 0 \quad \text{ on }\Gamma.
\end{aligned}
\right.$$
You can check the construction of yours $\langle xy,0,−y^2/2+xz \rangle $ is not divergence free, while formula (1) automatically gives you the divergence free potential field. For unbounded $\mathbb{R}^3$, choosing a gauge $g$ will get you the result you want.
I'll elaborate the idea of @RobBland .
Let us find $f$, so that $\nabla f=\mathbf{F}$. We will do that as follows:
$$
f(x,y)=\int P(x,y)dx=\int\left(1-\frac{x}{x^2+y^2}\right)dx=x-\frac{1}{2}\ln(x^2+y^2)+c(y).\tag 1
$$
We assume $y$ to be constant while performing integration $(1)$. $f(x,y)$ satisfies $\frac{\partial f}{\partial x}=P(x,y)$ condition if and only if it has the form $(1)$. Here $c(y)$ is some arbitrary (smooth enough) function; it depends only on $y$.
Now we should find $c(y)$ using $\frac{\partial f}{\partial y}=Q(x,y)$ condition. We will differentiate $(1)$ by $y$:
$$
\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\left(x-\frac{1}{2}\ln(x^2+y^2)+c(y)\right)=-\frac{y}{x^2+y^2}+c^\prime(y).\tag2
$$
But $\frac{\partial f}{\partial y}$ should equal $Q(x,y)=-\frac{y}{x^2+y^2}$:
$$
-\frac{y}{x^2+y^2}+c^\prime(y)=-\frac{y}{x^2+y^2}.
$$
Thus we obtain $c^\prime=0$, so $c=\mathrm{const}$. Now we got the potential:
$$
f(x,y)=x-\frac{1}{2}\ln(x^2+y^2)+c,
$$
here $c$ is an arbitrary constant, that may be omitted.
Now we obtain readily:
$$
\mathrm{Work}_{A\to B}=f(B)-f(A)=f(4,-3)-f(3,2)=\left[x-\frac{1}{2}\ln(x^2+y^2)\right]\Bigg|^{(4,-3)}_{(3,2)}=\\
=1-\frac{1}{2}\ln\frac{25}{13}
$$
REMARK
$f(x,y)$ may be found easier, if we recall, that $\left(-\frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2}\right)=-\frac{\mathbf{r}}{r^2}$ is the field of a charged straight wire (the wire is along $z$ axis). Its potential $f$ is $\ln r$ (up to some factor).
Best Answer
In order to determine a so that $\vec{f}$ is conservative, you must find:
$$ \oint \vec{F} \cdot d\vec{l} = 0 $$
$$ \int _{x_1} ^{x_2} F_x dx + \int _{y_1} ^{y_2} F_{y} dy + \int _{z_1} ^ {z_2} F_{z} dz = 0 $$
For the second question:
$$ \frac{\partial f}{\partial x} = F_x $$
$$ \frac{\partial f}{\partial y} = F_y $$
$$ \frac{\partial f}{\partial z} = F_z $$
Now you must solve the set of partial differential equation.