I'd try to use the following result:
$f$ has a removable singularity at $z=a$ if and only if $\lim_{z \to a} (z-a)f(z)=0$.
Since $f$ has a zero of order $m$ at $z=z_o$ it can be expressed as $f(z)=(z-z_o)^m \alpha (z)$; $g$ has a pole order $l$ at $z=z_o$ so it can be expressed as $g(z)=\frac{ \beta (z)}{(z-z_o)^l}$.
Then $f(z)g(z)=(z-z_o)^{m-l} \alpha(z) \beta (z)$, multiply by $z-z_o$, and then take a limit.
First question: A function is (complex) analytic if and only if it is holomorphic. It is an easy exercise to see that if $f$ is complex differentiable in $w$ and $f(w) \neq 0$, then $1/f$ is also complex differentiable in $w$ and $(1/f)'(w) = -f'(w)/f(w)^2$. So we know that $1/f$ is holomorphic in $B(z_0,\delta)\setminus \{z_0\}$, hence analytic.
Second question: It is not correct that $z_0$ is an isolated singularity of $1/f$ since $\lim\limits_{z\to z_0} \frac{1}{f(z)} = 0$. It is an isolated singularity since $1/f$ is holomorphic in $B(z_0,\delta)\setminus\{z_0\}$. The limit says that $z_0$ is a removable singularity, and the value which removes the singularity is $0$, so
$$g(z) = \begin{cases}\frac{1}{f(z)} &, z \neq z_0\\ 0 &, z = z_0 \end{cases}$$
is holomorphic on $B(z_0,\delta)$.
Third question: Since $g(z_0) = 0$, the power series representation of $g$ about $z_0$ has a constant term zero, hence
$$g(z) = \sum_{k=1}^\infty a_k (z-z_0)^k.$$
since $g$ does not vanish identically, not all coefficients are $0$, and if we let $m = \min \{ k \in\mathbb{N} : a_k \neq 0\}$, we have
$$g(z) = \sum_{k=m}^\infty a_k (z-z_0)^k$$
with $a_m \neq 0$. Then
$$h(z) = \sum_{k=m}^\infty a_k (z-z_0)^{k-m} = \sum_{r=0}^\infty a_{r+m}(z-z_0)^r$$
is holomorphic in $B(z_0,\delta)$ with $h(z_0) = a_m \neq 0$, and evidently $g(z) = (z-z_0)^m\cdot h(z)$. Since $g$ has no zero in $B(z_0,\delta)\setminus\{z_0\}$, $h$ does not vanish anywhere in $B(z_0,\delta)$, so $\tilde{h}(z) = \frac{1}{h(z)}$ is holomorphic on $B(z_0,\delta)$, and we have
$$f(z) = (z-z_0)^{-m}\cdot \tilde{h}(z)$$
on $B(z_0,\delta)\setminus\{z_0\}$.
Best Answer
If $z_0$ is removable, then $f$ admits a Laurent series representation in some punctured disc $D^*(z_0,r)$ with singular part equal to zero, so:
$$f(z)=\sum_{n=0}^{\infty} {a_n}(z-z_0)^n$$
Suppose the first $k-1$ terms of the Laurent series are zero, so $a_k$ is the first non zero term. Then:
$$f(z)=(z-z_0)^k\sum_{n=0}^{\infty} {a_{n+k}}(z-z_0)^n$$
We then have that: $$f(z)=(z-z_0)^k(a_k+a_{k+1}(z-z_0)+...) =(z-z_0)^kg(z)$$
With $g(z_0)=a_k\neq 0$ by hypothesis.