Shows that if $f(z)$ is a non-constant analytic function on a domain D, then the image under f(z) of any open set is an open set.
What I have so far: Since $f(z)$ is non-constant and and analytic, it is not identically zero and thus has isolated zeros. Likewise, since $f(z)$ is non-constant, $f'(z)$ is not identically zero so it has isolated zeros.
When $f'(z)$ is not equal to zero on some open ball $B_r(z_0) \in D$, where $D$ is a domain, then $|f'(z)|=J_f(z)$ does not equal zero and the inverse function theorem can be applied, mapping open balls to open balls.
Around isolated zeros of $f'(z)$ map open balls $B_\epsilon(a)$, where a is a zeros of $f'(z)$ by applying the following fact: if the analytic function $f(z)$ has a zero of order N at $z_0$, then $f(z)=g(z)^N$ for some function analytic near $z_0$ and satisfying $g'(z_0)$ does not equal zero.
Apparently this is supposed to reducing down to the fact that $h(z)=z^N$ maps open balls to open balls... something else I am not seeing.
If $z = a$ is an isolated zero of $f'(z)$, write $f$ as $$ f(z) = f(a) + (z - a)^N g(z) $$ where $N \in \mathbb{N}$ and $g$ is a holomorphic function with $g(a) \neq 0$. Since $g(a) \neq 0$, you can locally extract a holomorphic $N$-th root and obtain a holomorphic function $r$ defined near $z = a$, such that $r^N(z) = g(z)$. Then, if you define $h(z) = (z - a) r(z)$, you have near $z = a$ $$ f(z) = f(a) + h(z)^N $$ where $h'(a) = r(a) \neq 0$. Check that the map $z \mapsto z^N$ is open for all $N \in \mathbb{N}$. Then, near $z = a$, the map $f$ is a translation by $f(a)$ of a composition of two open maps (by what you wrote above), and thus an open map. Since it is enough to check that $f$ is open locally, we are done.