Answer
1. The work done by the field along the path $y = {x^2}$ is $W=1$.
2. The work done by the field along the path $x = {y^2}$ is $W=1$.
Work Step by Step
We have the field ${\bf{F}} = \left( {x + y,x - y} \right)$.
Case 1. The object moves from $\left( {0,0} \right)$ to $\left( {1,1} \right)$ along the path $y = {x^2}$.
The path can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,{t^2}} \right)$ for $0 \le t \le 1$.
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {t + {t^2},t - {t^2}} \right)$
${\bf{r}}'\left( t \right) = \left( {1,2t} \right)$
By Eq. (9), the work done by the field:
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^1 \left( {t + {t^2},t - {t^2}} \right)\cdot\left( {1,2t} \right){\rm{d}}t = \mathop \smallint \limits_0^1 \left( {t + 3{t^2} - 2{t^3}} \right){\rm{d}}t$
$W = \left( {\frac{1}{2}{t^2} + {t^3} - \frac{1}{2}{t^4}} \right)|_0^1 = \frac{1}{2} + 1 - \frac{1}{2} = 1$
So, the work done by the field along the path $y = {x^2}$ is $W=1$.
Case 2. The object moves from $\left( {0,0} \right)$ to $\left( {1,1} \right)$ along the path $x = {y^2}$.
The path can be parametrized by ${\bf{r}}\left( t \right) = \left( {{t^2},t} \right)$ for $0 \le t \le 1$.
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^2} + t,{t^2} - t} \right)$
${\bf{r}}'\left( t \right) = \left( {2t,1} \right)$
By Eq. (9), the work done by the field:
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^1 \left( {{t^2} + t,{t^2} - t} \right)\cdot\left( {2t,1} \right){\rm{d}}t = \mathop \smallint \limits_0^1 \left( {2{t^3} + 3{t^2} - t} \right){\rm{d}}t$
$W = \left( {\frac{1}{2}{t^4} + {t^3} - \frac{1}{2}{t^2}} \right)|_0^1 = \frac{1}{2} + 1 - \frac{1}{2} = 1$
So, the work done by the field along the path $x = {y^2}$ is $W=1$.