Why $f_x$ and $f_y$ should be continuous at z for differentiability theorem in complex analysis? Suppose$f_x$ and $f_y$ exist in a neighborhood of z. Then if $f_x$ and $f_y$ are continuous at z and $f_x$ =$if_y$ there, $f$ is differentiable (or holomorphic) at z.
Proof: Let  $f$ =  $u$ + i $v$, h = $\xi$ + i$\eta$. By the Mean-Value Theorem (for real functions of a real variable).
$\frac{u(z+h)-u(z)}{h}$= $\frac{u(x+\xi, y + \eta)-u(x,y)}{\xi + i\eta}$ = $\frac{u(x+\xi, y + \eta)-u(x+\xi,y)}{\xi + i\eta}$ + $\frac{u(x+\xi, y)-u(x,y)}{\xi + i\eta}$
=$\frac{\eta}{\xi+i\eta}u_y(x+\xi, y+\theta_1\eta) + \frac{\xi}{\xi+i\eta}u_x(x+\theta_2\xi,y) $
My professor said that I need $f_x$ and $f_y$ continuous at z in order to apply Mean-Value Theorem. Why do I need derivatives to be continuous and their existence is not enough? Thanks for your help.
 A: The version of the mean value theorem you are using establishes the existence of $c_1 \in (y,y+\eta)$ such that
$$  \frac{u(x+\xi, y+\eta) - u(x+\xi,y)}{\xi + \mathrm{i}\eta} = \frac{\eta}{\xi + \mathrm{i}\eta}u_y(x+\xi, c_1)  \text{.}  $$
This is not the form given.  To get $c_1 = y + \theta_1 \eta$ we must know that $u_y(x+\xi, y + \theta_1 \eta)$ takes every value in the set
$$  \left\{ u_y(x+\xi, c) \mid c \in (y,y+\eta) \right\}  \text{.}  $$
An easy way to do so is to apply the intermediate value theorem, which here requires that $u_y$ is continuous in its second argument.  That is, $u$ is continuously differentiable with respect to $y$.
The other term in your long equation give that $u$ is continuously differentiable with respect to $x$.  The same development for $v$ gives the requirement that it is continuously differentiable in each of $x$ and $y$.
One hypothesis that gives all four of these is that $f_x = u_x + \mathrm{i} v_x$ and $f_y = u_y + \mathrm{i} v_y$ are continuous, so $f$ is continuously differentiable with respect to both $x$ and $y$.
Maybe a weaker hypothesis would work.  $f$ merely differentiable in $x$ and $y$ is insufficient because there is no analog of the intermediate value theorem that can give the $\theta_1$ and $\theta_2$ for potentially nowhere continuous derivatives.  See this answer for a reminder of how awful the derivative of a differentiable but not continuous differentiable function can be -- and an easy visual example of why $\theta_1$ (and $\theta_2$) need not exist for such a function.
