Calculate the area of the region in the first quadrant... Calculate the area of the region in the first quadrant bounded by the
curves $4y = x, x^2y = 16, $ and $xy = 16.$

Was I right in adding the area under $y = 16/x^2$ from 8 to 4 back in?

You are making this more complicated than it needs to be. Why not just split the shaded region up into two regions, one from $x=1$ to $x=4$ and the other from $x=4$ to $x=8$. Then the area is
\begin{equation*}
  \int_1^4 \left(\frac{16}{x} - \frac{16}{x^2}\right)\,dx
      + \int_4^8 \left(\frac{16}{x} - \frac{x}{4}\right)\,dx.
\end{equation*}
