Where's this 1/2 coming from squaring the derivative for arc length? I am currently studying arc length in calculus 2 and I am having trouble understanding one of the solutions. I understand the concept and how the procedure works, but it seems like in a lot of problems you're adding on a 1/2 and bringing it in front of the integral, and I am not sure how or why.

Above is the image that shows the calculation for finding the derivative and then squaring that derivative. $1/2y^3 - 1/2y^{-3}$ squared should be $1/4y^6 + 1/4y^{-6}$.
Where is that $-1/2$ coming from in the 4th line of this solution? Is it so we can get a perfect square? And if so how do we just add that without changing the value of the equation?
Thanks!

You've forgotten the middle term in $(a-b)^2 = a^2 - 2 a b + b^2$ and that $y^3 \cdot y^{-3} = 1 \neq 0$.  Distibuting twice, then collecting...
\begin{align*}
 &\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right)^2  \\
  &\qquad = \left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right) \left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right)  \\
 &\qquad = \left( \frac{1}{2} \right) y^3\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right) - \left( \frac{1}{2} \right) y^{-3}\left( \left( \frac{1}{2} \right) y^3 - \left( \frac{1}{2} \right) y^{-3} \right)   \\
 &\qquad = \left( \left( \frac{1}{4} \right) y^6 - \left( \frac{1}{4} \right) y^{0} \right) - \left( \left( \frac{1}{4} \right) y^0 + \left( \frac{1}{4} \right) y^{-6} \right)  \\
 &\qquad = \left( \frac{1}{4} \right) y^6 - \left( \frac{2}{4} \right) y^{0} + \left( \frac{1}{4} \right) y^{-6}  \\
 &\qquad = \left( \frac{1}{4} \right) y^6 - \left( \frac{1}{2} \right) + \left( \frac{1}{4} \right) y^{-6}  \text{.}
\end{align*}
