It's usually a good idea to clear all the messy numbers and get them outside the integral out of the way so you can forget about them till the end. In this case, factorise the 16 out (take its square root first) and make it like:
Having done that, you can see a quadratic (disguised!) under that square root:
which I would be advised to investigate and see how it factorises.
In fact, to make your life far easier, I would have factorised the 3/4 out of the thing you needed to square further up, and got it completely out of the way.
This a good rule to learn. Factorise out the numbers whenever you can, because they just get in the way and interfere with the manipulations.
The quadratic does not factorise, instead you should notice
$\displaystyle \begin{align*} \sqrt{ \frac{ 9y^2 - 2y + 9}{y} } &= \frac{ \sqrt{ 9y^2 - 2y + 9 }}{\sqrt{y}} \\ &= \frac{ \sqrt{ 9 \left( \sqrt{y} \right) ^4 - 2\left( \sqrt{y} \right) ^2 + 9 } }{\sqrt{y}} \\ &= 2 \, \sqrt{ 9 \left( \sqrt{y} \right) ^4 - 2 \left( \sqrt{y} \right) ^2 + 9 } \, \left( \frac{1}{2\sqrt{y}} \right) \end{align*}$
and since $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}y} \left( \sqrt{y} \right) = \frac{1}{2\,\sqrt{y}} \end{align*}$, the substitution $\displaystyle \begin{align*} u = \sqrt{y} \implies \mathrm{d}u = \frac{1}{2\sqrt{y}}\,\mathrm{d}y \end{align*}$ is appropriate...