Amplitude transfer function
There are many algebraic paths to get into the required expression some of them being long and tedious, while the others being shorter with a carefully choosen route.

Given :
$$H(z) = \frac{ 1 - 2z^{-1} }{1 - z^{-1} + \frac{8}{9}z^{-2} }$$

To evaluate the corresponding Bode-Plot, by hand, we need to first derive the expression called as the Frequency Spectrum Magnitude $\lvert H(e^{j\omega})\rvert$ of $H(z)$, provided that the system is stable (i.e.all poles of $H(z)$ inside unit circle) And then compute:

$$20\log_{10}\left(\lvert H(e^{j\omega})\rvert\right),$$ 
to get the Magnitude in dB scale. 

Let's proceed into the first step and plug $e^{j\omega}$ into $H(z)$ to get Frequency Response:
$$H(e^{j\omega}) = \frac{ 1 - 2e^{-j\omega} }{1 - e^{-j\omega} + \frac{8}{9}e^{-2j\omega} }$$

As this is a complex valued rational fraction expression, it is equal to:
$$H(e^{j\omega}) = \frac{P(e^{j\omega})}{Q(e^{j\omega})}$$

And from complex algebra, the magnitude of $H(e^{j\omega})$ is equivalent to:
$$ \lvert H(e^{j\omega})\rvert = \frac{\lvert P(e^{j\omega})\rvert}{\lvert Q(e^{j\omega})\rvert}$$

Hence we need to compute the absolute values of the numerator and denumerator expressions, where the absolute value of a complex number is understood as: $$z=a+jb \implies \lvert z\rvert=(a^2+b^2)^{0.5}$$

Lets elaborate the Numerator:
\begin{align}
P(e^{j\omega}) &amp;= 1 - 2e^{-j\omega} = \left(1-2\cos(\omega)\right) + j\left(2\sin(\omega)\right)\\
\lvert P(e^{j\omega})\rvert &amp;= \left[\left(1-2\cos(\omega)\right)^2 + \left(2\sin(\omega)\right)^2\right]^{0.5}\\
\lvert P(e^{j\omega})\rvert &amp;= \left[1+4\cos^2(\omega)-4\cos(\omega) + 4\sin^2(\omega)\right]^{0.5}\\
\lvert P(e^{j\omega})\rvert &amp;= \left[5-4\cos(\omega)\right]^{0.5}\\
\end{align}

Also elaborate the Denumerator (Denominator I mean!):
\begin{align}
Q(e^{j\omega}) &amp;= 1 - e^{-j\omega} + \frac 89 e^{-2j\omega} = e^{-j\omega} \left(e^{j\omega} - 1 + \frac 89 e^{-j\omega}\right)\\
\lvert Q(e^{j\omega})\rvert&amp;=\lvert e^{-j\omega}\rvert\cdot \bigg\lvert \left(e^{j\omega} - 1 + \frac 89 e^{-j\omega}\right)\bigg\rvert\\
\lvert Q(e^{j\omega})\rvert&amp;=\bigg\lvert \left(e^{j\omega} - 1 + \frac 89 e^{-j\omega}\right)\bigg\rvert\\
\lvert Q(e^{j\omega})\rvert&amp;=\left[\left(\cos(\omega) - 1 + \frac 89 \cos(\omega)\right)^2 + \left(\sin(\omega) - \frac 89\sin(\omega)\right)^2\right]^{0.5}\\
\lvert Q(e^{j\omega})\rvert&amp;=\left[\left(\frac{17}{9}\cos(\omega) - 1\right)^2 + \left(\frac 19 \sin(\omega)\right)^2\right]^{0.5}\\
\end{align}

As there seems no more useful simplifications, we can deduce $\lvert H(e^{j\omega})\rvert$ as:
$$\lvert H(e^{j\omega})\lvert = \frac{\left[5-4\cos(\omega)\right]^{0.5}}{\left[\left(\frac{17}{9}\cos(\omega) - 1\right)^2 + \left(\frac 19\sin(\omega)\right)^2\right]^{0.5}}$$

The Bode plot takes the logarithm of this to produce the graph:
$$\begin{align} \lvert H(e^{j\omega})\rvert_\text{dB} &amp; \triangleq 20 \log_{10}\left(\lvert H(e^{j\omega})\rvert\right) \\
&amp; = 20 \log_{10}\left(\frac{\left[5-4\cos(\omega)\right]^{0.5}}{\left[\left(\frac{17}{9}\cos(\omega) - 1\right)^2 + \left(\frac 19\sin(\omega)\right)^2\right]^{0.5}}\right)\\ 
&amp; = 10 \log_{10}\left(\frac{5-4\cos(\omega)}{\left(\frac{17}{9}\cos(\omega) - 1\right)^2 + \left(\frac 19\sin(\omega)\right)^2}\right) \\ 
&amp; = 10 \log_{10}\left(5-4\cos(\omega)\right) - 10 \log_{10}\left(\left(\tfrac{17}{9}\cos(\omega) - 1\right)^2 + \left(\tfrac 19\sin(\omega)\right)^2\right)\\ 
\end{align}$$  

Which produces the same graph when compared with freqz([1 -2],[1 -1 8/9]) function of MATLAB to produce discrete time frequency response plot, for a linear frequency axis. In discrete time it is customary to use linear frequency axis rather than a logarithmic one as usual in continuous time Bode-Plots.