I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.
\begin{align}
&1 - \frac{-\frac{1}{3}\log_2\left(\frac{1}{3}\right) - \frac{1}{2}\log_2\left(\frac{2}{9}\right)}{-\sum\limits_{s = 1}^9 \frac{1}{9}\log_2\left(\frac{1}{9}\right)}\\\\
&\textrm{Denominator: You're just adding the same thing 9 times}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(\left(\frac{1}{3}\right)^{-1}\right) + \frac{1}{2}\log_2\left(\left(\frac{2}{9}\right)^{-1}\right)}{-9 \cdot \frac{1}{9}\log_2\left(\frac{1}{9}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{-\log_2\left(\frac{1}{9}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{\log_2\left(\left(\frac{1}{9}\right)^{-1}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{\log_2\left(9\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\left(\log_2\left(9\right) - \log_2\left(2\right)\right)}{\log_2\left(9\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(9\right) - \frac{1}{2}}{\log_2\left(9\right)}\\\\
\end{align}
For this to be equal to the second step, you'd need $\frac{1}{2}\log_2\left(9\right) - \frac{1}{2} = \frac{2}{9}\log_2\left(2\right) = \frac{2}{9}$ to be true. It's not: $\frac{1}{2}\log_2\left(9\right) - \frac{1}{2} \approx 1.1$ and $\frac{2}{9} \approx 0.2$.