Answer
The arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (\frac{1}{4}\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right),\frac{1}{{12}}{\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right)^{3/2}},$
${\ \ \ \ \ \ \ \ \ }$ $\frac{1}{{4\sqrt 3 }}{\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right)^{3/2}})$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,\frac{2}{3}{t^{3/2}},\frac{2}{{\sqrt 3 }}{t^{3/2}}} \right)$. So, the start point $\left( {0,0,0} \right)$ corresponds to $t=0$.
The derivative is
${\bf{r}}'\left( t \right) = \left( {1,{t^{1/2}},\sqrt 3 {t^{1/2}}} \right)$
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {1 + u + 3u} {\rm{d}}u = \mathop \smallint \limits_0^t \sqrt {1 + 4u} {\rm{d}}u$
Write $v=1+4u$. So, $dv = 4du$. The integral becomes
$s\left( t \right) = \frac{1}{4}\mathop \smallint \limits_1^{1 + 4t} {v^{1/2}}{\rm{d}}v = \frac{1}{4}\cdot\frac{2}{3}{v^{3/2}}|_1^{1 + 4t}$
$s\left( t \right) = \frac{1}{6}\left( {{{\left( {1 + 4t} \right)}^{3/2}} - 1} \right)$
It follows that
$6s + 1 = {\left( {1 + 4t} \right)^{3/2}}$
${\left( {6s + 1} \right)^{2/3}} = 1 + 4t$
$4t = {\left( {6s + 1} \right)^{2/3}} - 1$
The inverse of $s\left( t \right)$ is $t = \frac{1}{4}\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right)$. Since the start point $\left( {0,0,0} \right)$ corresponds to $t=0$, so at the start point we have $s=0$.
Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {t,\frac{2}{3}{t^{3/2}},\frac{2}{{\sqrt 3 }}{t^{3/2}}} \right)$ gives the arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (\frac{1}{4}\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right),\frac{1}{{12}}{\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right)^{3/2}},$
${\ \ \ \ \ }$ $\frac{1}{{4\sqrt 3 }}{\left( {{{\left( {6s + 1} \right)}^{2/3}} - 1} \right)^{3/2}})$
