#### Answer

$$2\sqrt{3}$$

#### Work Step by Step

Since
\begin{aligned}
1+\left(y^{\prime}\right)^{2} &=1+\left(\frac{(x-3)(x-1)}{6 y}\right)^{2} \\
&=1+\frac{(x-3)^{2}(x-1)^{2}}{36 y^{2}} \\
&=1+\frac{(x-3)^{2}(x-1)^{2}}{4\left(9 y^{2}\right)}\\
&= 1+\frac{(x-1)^{2}}{4 x}
\end{aligned}
Then
\begin{aligned}
s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x &=\int_{0}^{3} \sqrt{1+\frac{(x-1)^{2}}{4 x}} d x \\
&=\frac{1}{2} \int_{0}^{3} \sqrt{4+\frac{(x-1)^{2}}{x}} d x \\
&=\frac{1}{2} \int_{0}^{3} \frac{x+1}{\sqrt{x}} d x \\
&=\frac{1}{2} \int_{0}^{3} \sqrt{x}+\frac{1}{\sqrt{x}} d x \\
&=\frac{1}{2}\left[\frac{2}{3} x^{3 / 2}+2 \sqrt{x}\right]_{0}^{3}\\
&=2\sqrt{3}
\end{aligned}