Answer
The vector fields are ${\bf{F}} = \left( {g\left( y \right),h\left( x \right)} \right) = \left( {ay + 1,ax + 1} \right)$,
where $a$ is a constant.
Work Step by Step
We have a conservative vector field of the form ${\bf{F}}\left( {x,y} \right) = \left( {g\left( y \right),h\left( x \right)} \right)$.
Since ${\bf{F}}\left( {0,0} \right) = \left( {1,1} \right)$, so $g\left( 0 \right) = 1$ and $h\left( 0 \right) = 1$.
Write ${\bf{F}} = \left( {g\left( y \right),h\left( x \right),0} \right)$. Let ${\bf{F}}$ be a conservative vector field.
By Theorem 4 in Section 17.3, ${\bf{F}}$ satisfies the cross-partials condition:
$\frac{{\partial {F_1}}}{{\partial y}} = \frac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ \ }$ $\frac{{\partial {F_2}}}{{\partial z}} = \frac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ \ }$ $\frac{{\partial {F_3}}}{{\partial x}} = \frac{{\partial {F_1}}}{{\partial z}}$
That is,
$g'\left( y \right) = h'\left( x \right)$, ${\ \ \ \ }$ $0=0$, ${\ \ \ \ }$ $0=0$
Case 1. Integrate the differential equation with respect to $x$, gives
$\mathop \smallint \limits_{}^{} g'\left( y \right){\rm{d}}x = \mathop \smallint \limits_{}^{} h'\left( x \right){\rm{d}}x$
(1) ${\ \ \ \ \ \ \ }$ $x g'\left( y \right) = h\left( x \right) + {C_1}$
At $\left( {x,y} \right) = \left( {0,0} \right)$, we have $h\left( 0 \right) = 1$, so
$0 = 1 + {C_1}$
${C_1} = - 1$
Equation (1) becomes
$xg'\left( y \right) = h\left( x \right) - 1$
$h\left( x \right) = xg'\left( y \right) + 1$
Since $g'\left( y \right)$ is a function of $y$ only, whereas $h\left( x \right)$ is a function of $x$ only, we conclude that $g'\left( y \right)$ must be a constant. Thus
$h\left( x \right) = ax + 1$
where $a$ is a constant.
Case 2. Integrate the differential equation with respect to $y$, gives
$\mathop \smallint \limits_{}^{} g'\left( y \right){\rm{d}}y = \mathop \smallint \limits_{}^{} h'\left( x \right){\rm{d}}y$
(2) ${\ \ \ \ \ \ \ }$ $g\left( y \right) = yh'\left( x \right) + {C_2}$
At $\left( {x,y} \right) = \left( {0,0} \right)$, we have $g\left( 0 \right) = 1$, so
$1 = {C_2}$
Equation (2) becomes
$g\left( y \right) = yh'\left( x \right) + 1$
Since $h'\left( x \right)$ is a function of $x$ only, whereas $g\left( y \right)$ is a function of $y$ only, we conclude that $h'\left( x \right)$ must be a constant. Thus
$g\left( y \right) = by + 1$
where $b$ is a constant.
Since $g'\left( y \right) = h'\left( x \right)$, $g'\left( y \right) = b$ and $h'\left( x \right) = a$, so
$a=b$
Thus, $h\left( x \right) = ax + 1$ and $g\left( y \right) = ay + 1$.
The vector field is ${\bf{F}} = \left( {g\left( y \right),h\left( x \right)} \right) = \left( {ay + 1,ax + 1} \right)$.
