Calculating the quotient of 2 Laurent series

Let $D$ be an open subset of $\mathbb{C}$ and $f,g : D \rightarrow \mathbb{C}$ be meromorphic, both having a pole at $z_0$. If I know the Laurent series of $f$ and $g$ around $z_0$, how to calculate the Laurent series of $\frac{f}{g}$ around $z_0$? I made up an example and couldn't transform it into the standard form to see the coefficients. This is my example with $z_0=0$: $$f(z)=z^{-2}+3z+2z^2$$ $$g(z)=z^{-2}+4z+8z^3$$

Further, if $f$ and $g$ have the same principal part (like in my example), then I want to show that $\frac{f}{g}$ doesn't have a principal part. How can I show that?

Your second question sounds like a use of Morera's Theorem: If a function is analytic on $D$ except a single point $z_0$ where it is continuous, then it is analytic at that point as well.
So, $f/g$ won't have a principal part at $z_0$ because (with the assumption that they have the same pole of the same order) then the following limit will exist, $$\lim_{z\to z_0} \frac{f}{g} .$$ Like in your example, $$\lim_{z\to 0} \frac{f}{g} =1.$$
Your fist question, about how to find the Lauren series of a quotient, is much more difficult to answer. Typically, special cases are known and techniques are learned, but a general form might prove quite tedious. If you aren't allowing essential singularities, a fairly simple technique can be developed, let $$f(z)=\sum_{k=-n}^\infty a_k (z-z_0)^k\ \ \ \text{and}\ \ \ g(z)=\sum_{k=-m}^\infty b_k (z-z_0)^k,$$ where $a_{-n}$ and $b_{-m}$ are not zero. Then $$(z-z_0)^nf(z)\ \ \ \text{and}\ \ \ (z-z_0)^mg(z)$$ are analytic functions in a neighborhood of $z_0$, that are not zero at $z_0$. Hence, their quotient is an analytic function at $z_0$ and thus a taylor series can be found, like, $$\frac{f(z)}{g(z)}=(z-z_0)^{m-n} \left(\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right) = (z-z_0)^{m-n} \sum_{k=0}^\infty \left(\left.\frac{d^k}{dz^k}\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right|_{z=z_0}\right)(z-z_0)^k.$$ Hence, $$\frac{f(z)}{g(z)} = \sum_{k=0}^\infty \left(\left.\frac{d^k}{dz^k}\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right|_{z=z_0}\right)(z-z_0)^{m-n+k}.$$ So, if you were asked in integrate this, you would simply need to find the $k$ where $m-n+k=-1$ and then use the coefficient at that part and calculate, $$\int_C \frac{f(z)}{g(z)}dz = 2\pi i \cdot \left.\frac{d^{n-m-1}}{dz^{n-m-1}}\frac{(z-z_0)^nf(z)}{(z-z_0)^mg(z)}\right|_{z=z_0}.$$ If you actually calculate this out for specific $n$ and $m$ then you obtain many residue formulas found in textbooks. Take a look at Marsden - Complex Analysis - Page 250, which presents a table dedicated to this very concept.