Question: 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?

Answer: 
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 $\omega$