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
The function $$g(z) = \frac{f(z) - 1}{f(z) + 1}$$ has an isolated singularity at $a$ and satisfies $|g(z)| < 1$ in a (deleted) neighborhood of $a$, thus $\lim_{z\to a} g(z)$ exists. Hence $$\lim_{z\to a} (z - a)f(z) = \lim_{z\to a} (z - a)\frac{1 + g(z)}{1 - g(z)} = \lim_{z\to a} (z - a)(1 + g(z) + O(g(z)^2)) = 0.$$ Consequently, $a$ is a removable singularity of $f$.