Personally, I would approach the problem by parametrizing the curve as follows:
$$ x = t^2 - 3 \qquad \qquad y = 4t^3 \qquad (\sqrt{3} \leq t \leq \sqrt{6})$$
In which case we see that:
$$ \frac{dx}{dt} = 2t \qquad \qquad \frac{dy}{dt} = 12t^2 $$
and so the arc length is expressed as:
\begin{align*}
L &= \int_C \sqrt{\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2} \, dt \\
&= \int_\sqrt{3}^\sqrt{6} \sqrt{(2t)^2 + (12t^2)^2} \, dt \\
&= \int_\sqrt{3}^\sqrt{6} \sqrt{144t^4 + 4t^2} \, dt \\
&= \int_\sqrt{3}^\sqrt{6} 2t \sqrt{36t^2 + 1} \, dt \\
&= \int_3^6 \sqrt{36u + 1} \, du \\
&= \left[ \frac{1}{54}(36u+1)^\frac{3}{2} \right]_3^6 \\
&= \frac{217\sqrt{217} - 109\sqrt{109}}{54} \approx 38.1225
\end{align*}
Though I come to see that the problem also works out nicely in Cartesian coordinates rather than parametric coordinates:
\begin{align*}
y &= 4(x+3)^{3/2} \\
\frac{dy}{dx} &= 6\sqrt{x+3}
\end{align*}
So the arc length is expressed as:
\begin{align*}
L &= \int_C \sqrt{1 + \left(\frac{dy}{dx} \right)^2} \, dx \\
&= \int_0^3 \sqrt{1 + 36(3+x)} \, dx \\
&= \int_0^3 \sqrt{36x + 109} \, dx \\
&= \left[\frac{1}{54}(36x+109)^\frac{3}{2} \right]_0^3 \\
&= \frac{217\sqrt{217} - 109\sqrt{109}}{54} \approx 38.1225
\end{align*}