Three. (25 points) Find the minimum value of the function
$$
y=\sqrt{x^{4}-3 x^{2}-6 x+13}+\sqrt{x^{4}-x^{2}+1}
$$
$$
\begin{aligned}
y= & \sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}}+ \\
& \sqrt{(x-0)^{2}+\left(x^{2}-1\right)^{2}} .
\end{aligned}
$$

The above expression represents the sum of the distances from point $P\left(x, x^{2}\right)$ to points $A(3,2)$ and $B(0,1)$. The image of point $P\left(x, x^{2}\right)$ is the parabola $y=x^{2}$. As shown in Figure 5, connecting $A B$ intersects the parabola $y=x^{2}$ at point $P$. Thus, the minimum value of $y$ is the length of line segment $A B$:
$$
|A B|=\sqrt{(3-0)^{2}+(2-1)^{2}}=\sqrt{10} \text {. }
$$

It is easy to find that the equation of line $A B$ is $y=\frac{1}{3} x+1$.
By solving $\left\{\begin{array}{l}y=x^{2}, \\ y=\frac{1}{3} x+1\end{array}\right.$
$\Rightarrow\left\{\begin{array}{l}x=\frac{1+\sqrt{37}}{6}, \\ y=\frac{19+\sqrt{37}}{18}\end{array}\right.$ (negative value has been discarded).
Therefore, when $x=\frac{1+\sqrt{37}}{6}$, the minimum value of $y$ is $\sqrt{10}$.