The area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x$-$y$ plane is $\frac{1}{3}$.
To find the area, we can use the following formula:
$$\text{Area}=\int_{a}^{b}\left(f(x)-g(x)\right)\,dx$$
where $f(x)$ and $g(x)$ are the functions that define the boundaries of the area, and $a$ and $b$ are the $x$-coordinates of the endpoints of the area.
In this case, $f(x)=x$ and $g(x)=x^2$, so we have:
$$\text{Area}=\int_{0}^{1}\left(x-x^2\right)\,dx=\int_{0}^{1}-x^2\,dx$$
We can evaluate this integral using the following formula:
$$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$
where $C$ is an arbitrary constant.
In this case, $n=2$, so we have:
$$\begin{align}
\text{Area}&=\int_{0}^{1}-x^2\,dx \\
&=-\frac{x^3}{3}~\Bigg|_{0}^{1} \\
&=-\frac{1}{3}-\left(0\right) \\
&=\frac{1}{3}
\end{align}$$
Therefore, the area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x$-$y$ plane is $\frac{1}{3}$.