Proving that this function is continuous on $G\times G$

On the open set $\Omega = G \setminus \Delta_G$, both $z - w$ and $f(z) - f(w)$ are continuous and $z - w$ has no zeros, so $g(z,w)$ is continuous. So we only need to show that $g(z,w)$ is continuous at points in $\Delta_G$, i.e. $$\Vert (z,w) - (z_0,z_0) \Vert < \delta \implies \vert g(z,w) - g(z_0,z_0) \vert < \epsilon.$$

If $z = w$ this is easy, as then $$g(z,w) - g(z_0,z_0) = f'(z) - f'(z_0)$$ and $f'$ is continuous. If $z \neq w$, we have $$g(z,w) - g(z_0,z_0) = \frac{f(z) - f(w)}{z - w} - f'(z_0).$$ For this case write $$f(z) = \sum_{n=0}^{\infty}a(n)(z - z_0)^n.$$ Then \begin{equation*} \begin{aligned} \frac{f(z) - f(w)}{z - w} &= \frac{1}{z - w}\sum_{n=0}^{\infty}a(n)[(z - z_0)^n - (w - z_0)^n] \\ &= \frac{1}{(z - z_0) - (w - z_0)}\sum_{n=1}^{\infty}a(n)[(z - z_0)^n + (w - z_0)^n] \\ &= \sum_{n=1}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}] \\ \end{aligned} \end{equation*} with $Z = z - z_0, W = w - z_0$. We have $$\frac{f(z) - f(w)}{z - w} - f'(z_0) = \sum_{n=2}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}].$$

Thus if $\vert Z \vert, \vert W \vert < \delta$, then $$\left\vert \frac{f(z) - f(w)}{z - w} - f'(z_0) \right\vert < \sum_{n=2}^{\infty}n \vert a(n) \vert\delta^{n - 1} < \epsilon.$$