fractional linear transformations 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.

 A: 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.
A: There are several ways to do these, but most of them are facilitated by sort of knowing the answer. Recall that fractilinear transformations preserve circilinearity, i.e. circles and lines get mapped to circles and lines (NOT respectively - perhaps a circle to a line or vice versa). And a line is a circle that goes through infinity (really - it's a good way to think of these things).
So if I were you, I would get an idea for where the circle goes by looking at the boundaries of the complex unit cube, or perhaps the four easy coordinates (the purely real and the purely imaginary) of the unit circle (or both), and see where they go. This gives a good idea.
Also, I would note that the new 'infinity point' is when the denominator is zero.
Does that give you a good start?
A: If you plug in $z = e^{i\theta}$ into your linear fractional transformation you get
$${3e^{i\theta} + i \over -ie^{i\theta} + 3}$$
This is the same as 
$$e^{i\theta} {3 + ie^{-i\theta} \over 3 - ie^{i\theta}}$$
The numerator and denominators of the fraction are complex conjugates of each other, so it has magnitude $1$, as does $e^{i\theta}$. So their product also has norm $1$ for all $\theta$ and therefore the linear fractional transformation takes the unit circle to itself.
In general when it takes the unit circle to itself you can use factorizations this way to show it. 
