I can prove that the function is continuous on $(z,w)$ where $z\neq w$ by showing for any sequence $(z_n,w_n)\rightarrow (z,w)$,
$lim $ $\phi((z_n,w_n))\rightarrow\phi((z,w))$
I can do this since $\exists N\forall n\geq n (Z_n\neq W_n)$
Hence, $lim $ $\phi((z_n,w_n))=lim \frac{f(z_n)-f(w_n)}{z_n-w_n}= \frac{f(z)-f(w)}{z-w}$
However, Im stuck at showing continuity when z=w since the function is not always of the same form. I can't seem to provide epsilon proof as well.
Any solutions?
If $(z_n,w_n) \to (z,z)$ then $z_n \to z$ and $w_n \to z$. Assume first that $z_n\neq w_n$ for all $n$. Now $\frac {f(z_n)-f(w_n)} {z_n-w_n}=\frac 1 {z_n-w_n} \int_L f'(\zeta) \, d\zeta$ where $L$ is the line segment from $z_n$ to $w_n$. Continuity of $f'$ makes it easy to see that $\frac {f(z_n)-f(w_n)} {z_n-w_n} \to f'(z)$: $|\frac 1 {z_n-w_n} \int_L f'(\zeta) \, d\zeta - f'(z)|=|\frac 1 {z_n-w_n} \int_L \{f'(\zeta)-f'(z)\} \, d\zeta $. Given $\epsilon >0$ choose $\delta >0$ such that $|f'(\zeta)-f'(z)|<\epsilon$ if $|\zeta -z| <\delta$. Note that if $n$ is sufficiently large then $|\zeta -z| <\delta$ for all points $\zeta$ on the line $L$. Hence $|\frac 1 {z_n-w_n} \int_L f'(\zeta) \, d\zeta - f'(z)| <\epsilon$ for $n$ large enough. Now if $z_n=w_n$ for all $n$ then $\phi (z_n,w_n)=f'(z_n) \to f'(z)=\phi (z,z)$. Finally any sequence $(z_n,w_n)$ converging to $(z,z)$ can be split into (at most) two subsequences of the two types we have considered.