Answer
The potential function does not exist.
Work Step by Step
Suppose that $f$ is a potential function for ${\bf{F}}$. Then, $\nabla f = {\bf{F}} = \left( {yz,xz,y} \right)$.
Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$, so
(1) ${\ \ \ \ \ }$ $\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {yz,xz,y} \right)$
Consider $\frac{{\partial f}}{{\partial x}} = yz$.
Integrating with respect to $x$ gives
$f\left( {x,y,z} \right) = xyz + g\left( {y,z} \right)$,
where $g$ is a function of $y$ and $z$.
Taking the derivatives of $f\left( {x,y,z} \right)$ with respect to $z$ gives $\frac{{\partial f}}{{\partial z}} = xy + \frac{{\partial g}}{{\partial z}}$. But this is not $\frac{{\partial f}}{{\partial z}} = y$ as required. So, we conclude that there is no function $f$ such that equation (1) is satisfied, so the potential function does not exist.
