Answer
By inspection, a potential function for ${\bf{F}}$ is $f = \frac{1}{2}{x^2}$.
We prove that ${\bf{G}} = \left( {y,0} \right)$ is not conservative.
Work Step by Step
1. We have the vector field: ${\bf{F}} = \left( {x,0} \right)$.
Let $f$ be a potential function for ${\bf{F}}$. Then, $\nabla f = {\bf{F}} = \left( {x,0} \right)$.
Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right)$, so
$\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) = \left( {x,0} \right)$
$\frac{{\partial f}}{{\partial x}} = x$, ${\ \ \ \ \ }$ $\frac{{\partial f}}{{\partial y}} = 0$
So, $f = \frac{1}{2}{x^2}$.
2. We have the vector field: ${\bf{G}} = \left( {y,0} \right)$.
Write ${\bf{G}} = \left( {{G_1},{G_2}} \right) = \left( {y,0} \right)$. Evaluate:
$\frac{{\partial {G_1}}}{{\partial y}} = 1$, ${\ \ \ \ \ }$ $\frac{{\partial {G_2}}}{{\partial x}} = 0$
Since $\frac{{\partial {G_1}}}{{\partial y}} \ne \frac{{\partial {G_2}}}{{\partial x}}$, by Theorem 1, ${\bf{G}} = \left( {y,0} \right)$ is not conservative.
