ok bit of a silly question but i really dont understand how to rearange
$\displaystyle \frac{\sqrt{2}}{2\sqrt{y}}$
to get
$\displaystyle \frac{1}{\sqrt{2y}}$
No, you can treat this sum as $\displaystyle \frac{\sqrt2}{2} \times \frac{1}{\sqrt{y}}$
As $\displaystyle \frac{\sqrt2}{2} = \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}} = \frac{1}{\sqrt2}$ then it becomes $\displaystyle \frac{1}{\sqrt2 \sqrt{y}}$
As $\displaystyle \sqrt{a} \sqrt{b} = \sqrt{ab} \: , \: a,b \geq 0$ we can combine the square roots in the denominator:
$\displaystyle \frac{1}{\sqrt{2y}}$