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So I have the following equation $$ y[n]=0.9y[n-1]+0.1x[n] $$
We can find easily find the transfer function, which is
$$ H(z)=\frac{0.1}{1-0.9z^{-1}} $$
and from that, the frequency response, which is
$$ H(e^{j \omega})=\frac{0.1}{1-0.9 e^{-j \omega}} $$
To find the cutoff frequency we just make the magnitude of the above equal to $\frac{1}{\sqrt{2}}$. Then, $\omega_0 = 0.0675\Rightarrow f_0 = 0.0107$.
Okay, so how can we say that given $f_0 = f_c$ the filter is a LPF or HPF?
Types of filters, such as LPF, HPF, BPF or BSF, are not described by their cutoff frequencies $f_c$ but through their passband or stopband frequency $\omega$ region.
For example, in your case of the filter $H(z) = 1 / (1 - 0.9 z^{-1})$, the passband is in the low frequencies (where the magnitude of the frequency response $H(e^{j\omega})$ is greater than $1/\sqrt{2}$. And by your definition that region corresponds to $\omega \in (-f_c , f_c)$ which indicates a lowpass range of frequencies.
Other filters like HPF and BPF follow in a similar manner. Note that specific filters such as comb or notch types are defined in a different style than this.
It's actually the Z Transform
\begin{equation} H(z) = \frac{0.1}{1-0.9z^{-1}} \end{equation}
Now get the frequency response by evaluating \begin{equation} H(e^{j\omega}) = \frac{0.1}{1-0.9z^{-1}} \Bigr|_{z = e^{j\omega}} \end{equation}
You can now plot magnitude and phase for a given frequency $\omega$
