From my research, I have figured out that this is a Möbius transformation. The respective wiki page helped me understand a bit more, however I can't figure out how to obtain the image.
So lets talk about what I do know. Well we are describing the set of all values $f(z)$ where $|z| < 1$. I also know that any three points.
I also know that
Given a set of three distinct points z1, z2, z3 on the Riemann sphere and a second set of distinct points w1, w2, w3, there exists precisely one Möbius transformation f(z) which maps the zs to the ws
and from help from another forum I got the following:
To decide which part is the image of the interior |z| < 1 of the disc,
figure out which point f sends to infinity.
Best Answer
Hint. The inverse of $w=\dfrac{3z+i}{-iz+3}$ is $z=\dfrac{3w-i}{iw+3}$. So $|z|<1$ is equivalent to $|3w-i|<|iw+3|$, or $3|w-\dfrac{i}{3}|<|i||w-3i|=|w-3i|$. One can find the desired image by considering the Apollonius' circle in the complex plane ($w$-plane) or by squaring both sides and using $|z|^2=z\overline{z}$.
Edit: To elaborate on the 'square both sides and use $|z|^2=z\overline{z}$' part: $$\begin{align*} |3w-i|^2&<|iw+3|^2\\ (3w-i)(3\overline{w}+i)&<(iw+3)(-i\overline{w}+3)\\ 9w\overline{w}+3iw-3i\overline{w}+1&<w\overline{w}+3iw-3i\overline{w}+9\\ 8w\overline{w}&<8\\ |w|^2&<1\\ |w|&<1 \end{align*}$$
Hence the image is again the open unit disk.