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!
Usually the definition of a conservative vector field is this:
Let $\vec{f}$ be a vector field in an open set $\Omega\subset\mathbb R^2$.
It is called conservative if there is a scalar function $\Phi:\Omega\to\mathbb R$ so that $\vec f=\vec\nabla\Phi$.
This definition makes no assumptions about the geometry.
Then we have these two theorems:
- If $f$ is conservative, then $\partial_1f_2=\partial_2f_1$.
- If $\Omega$ is simply connected, then $\partial_1f_2=\partial_2f_1$ implies that $f$ is conservative.
Star-shaped domains are simply connected, but the assumption is more general than that.
Using the first theorem we can see that since $f$ fails the conclusion, it must also fail the assumption; that is, $f$ is not conservative.
The special thing about simply connected domains is that they have no holes: you cannot take a path around a point outside $\Omega$.
It is the existence of such paths that causes theorem 2 to fail if $\Omega$ is not simply connected.
(There is also a measure for how badly it fails for a given domain, namely the first de Rham cohomology group.)
Star-shaped domains are nice because you can find a very simple path between any two points (through the center).
In more general simply connected domains you need more complicated paths but the basic idea is the same.
(Similar ideas work in higher dimensions as well, but I restricted myself to the plane for simplicity.)
Best Answer
Since $F_i=kr^{-3}x_i$, $$\partial_iF_j=k(r^{-3}\delta_{ij}-3r^{-4}x_j\partial_i r)=k(r^{-3}\delta_{ij}-3 r^{-5}x_i x_j)$$is $i\leftrightarrow j$ symmetric, so$$(\nabla\times F)_h=\sum_{ij}\epsilon_{hij}\partial_i F_j=0.$$