Figuring magnitude and phase response
$$
|H(z)| = \sqrt{\Re^2 H(z) + \Im^2 H(z)} \\
\theta \big(H(z)\big) = \arctan \frac{\Im H(z)}{\Re H(z)} \\
$$

$\Re z$ is the real part of z;

$\Im z$ is imaginary part.

For a single-pole filter $y_n = b_0 x_n - a_1 y_{n-1}$,

$$
H(z) = \frac{b_0}{1 + a_1 z^{-1}}; \\
\Re H(z) = b_0 \frac{1 + a_1 \cos \omega}{1 + 2 a_1 \cos \omega + a_1^2}; \\
\Im H(z) = b_0 \frac{a_1 \sin \omega}{1 + 2 a_1 \cos \omega + a_1^2}.
$$

With $b_0 = 1$ and $a_1 = -\frac{8}{9}$, your filter may be unstable.

For a low-pass filter, try $b_0 = \frac{1}{9}, a_1 = -\frac{8}{9}$.