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?