I am currently refreshing my complex analysis knowledge (I had a course many years ago, but I've forgotten almost all of it). The German textbook I'm using, which is Funktionentheorie by Fischer & Lieb, states the following definition of real differentiability
A function $g:U\to\Bbb R$ is said to be real differentiable in the point $z_0=x_0+iy_0$ of the domain $U\subseteq\Bbb C$ if there exist real-valued functions $\Delta_1$ and $\Delta_2$ on $U$ which are continuous at $z_0$ such that for all $z=x+iy\in U$ we have $$ g(z) = g(z_0) + \Delta_1(z)(x-x_0) + \Delta_2(z)(y-y_0) \qquad\qquad (1) $$
I found myself wondering how this is related to the usual definition of a partially differentiable function.
If some function $g:U\to\Bbb R$ satisfies this condition, and we define $\Delta_1^{y_0}(x)=\Delta_1(x,y_0)$ and $\Delta_2^{x_0} = \Delta_2(x_0,-)$, then
$$
g(x,y_0) = g(x_0,y_0) + \Delta_1^{y_0}(x)(x-x_0) \quad\text{and}\quad
g(x_0,y) = g(x_0,y_0) + \Delta_2^{x_0}(y)(y-y_0)
$$
so $\Delta_1^{y_0}(x_0) = \frac{\partial g}{\partial x}(z_0) = \lim_{x\to x_0}\frac{g(x,y_0)-g(x_0,y_0)}{x-x_0}$ and $\Delta_2^{x_0}(z_0) = \frac{\partial g}{\partial y}(z_0) = \lim_{y\to y_0}\frac{g(x_0,y)-g(x_0,y_0)}{y-y_0}$.
So the partial derivatives exist. Moreover, $g$ is continuous at $z_0$.
Now assume for the other direction that $g$ is continuous at $z_0$ and that $\Delta_1^{y_0}$ and $\Delta_2^{x_0}$ exist as above (so they are continuous at $x_0$ and $y_0$, respectively). I figured out that if we set $$ \Delta_1(x,y) = \frac12\left(\Delta_1^{y_0}(x) + \frac{g(x,y)-g(x_0,y)}{x-x_0}\right) \\ \Delta_2(x,y) = \frac12\left(\Delta_2^{x_0}(y) + \frac{g(x,y)-g(x,y_0)}{y-y_0}\right) $$ then this satisfies (1). Note that we let the right-hand summand in the first line assume the value $\Delta_1^{y_0}(x_0)$ at $(x_0,y_0)$, and an arbitrary value at any other $(x_0,y)$, and similarly for the summand in then second line.
Now my problem: I have trouble showing that the right-hand summand in the first line is continuous at $(x_0,y_0)$. Any ideas? Or maybe definition above is not equivalent to the existence of the partial derivatives? Is it even equivalent to anything else?
Some words about the context: In the textbook, three pages later, it is shown that a function $f=g+ih:U\to\Bbb C$ is complex differentiable in $z_0$ if and only if $\frac{\partial f}{\partial \overline z}(z_0)=0$ and the functions $g$ and $h$ are real differentiable in the sense of (1).
The definition captures total differentiability, not partial differentiability. It is directly taken from the definition of differentiability in Grauert/Fischer, Differential- und Integralrechnung II, chapter 3, on page 50, with the difference that in Grauert/Fischer, the domain of the function is not required to be open.
You are probably more used to total differentiability of $g$ at $z_0$ being characterised by the existence of a real-linear map $L \colon \mathbb{C}\to \mathbb{R}$ such that
$$g(z) - g(z_0) - L(z-z_0) \in o(\lvert z-z_0\rvert).\tag{2}$$
The existence of a representation $(1)$ implies $(2)$, by setting $L(w) = \Delta_1(z_0)\cdot u + \Delta_2(z_0)\cdot v$, with $u = \operatorname{Re} w$, $v = \operatorname{Im} w$, since by continuity of $\Delta_k$ at $z_0$, for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\lvert \Delta_k(z) - \Delta_k(z_0)\rvert \leqslant \varepsilon$ for $\lvert z-z_0\rvert \leqslant \delta$, and thus
\begin{align} \lvert g(z) - g(z_0) - L(z-z_0)\rvert &= \bigl\lvert\bigl(\Delta_1(z) - \Delta_1(z_0)\bigr)(x-x_0) + \bigl(\Delta_2(z) - \Delta_2(z_0)\bigr)(y-y_0)\bigr\rvert\\ &\leqslant \lvert\Delta_1(z) - \Delta_1(z_0)\rvert\cdot\lvert x-x_0\rvert + \lvert \Delta_2(z) - \Delta_2(z_0)\rvert\cdot\lvert y-y_0\rvert\\ &\leqslant \varepsilon\cdot(\lvert x-x_0\rvert + \lvert y-y_0\rvert)\\ &\leqslant \varepsilon\sqrt{2}\cdot \lvert z-z_0\rvert. \end{align}
Conversely, if we have $(2)$, we can define
$$\mathrm{E}(z) = \begin{cases}\dfrac{g(z) - g(z_0) - L(z-z_0)}{z - z_0} &, z \neq z_0 \\ \qquad\qquad 0 &, z = z_0 \end{cases},$$
which is continuous at $z_0$ by $(2)$, and set
\begin{align} \tilde{\Delta}_1(z) &= \frac{\partial g}{\partial x}(z_0) + \mathrm{E}(z),\\ \tilde{\Delta}_2(z) &= \frac{\partial g}{\partial y}(z_0) + i\mathrm{E}(z), \end{align}
using $L(z-z_0) = \frac{\partial g}{\partial x}(z_0)\cdot (x-x_0) + \frac{\partial g}{\partial y}(z_0)\cdot (y-y_0)$. Then $\tilde{\Delta}_k$ is continuous at $z_0$, and we have the representation
$$g(z) = g(z_0) + \tilde{\Delta}_1(z)(x-x_0) + \tilde{\Delta}_2(z)(y-y_0).$$
That's almost the representation $(1)$, only the $\tilde{\Delta}_k$ are in general not real-valued. Small problem, taking $\Delta_k(z) = \operatorname{Re} \tilde{\Delta}_k(z)$ gives us $(1)$.
Thus we have seen that $(1)$ is equivalent to the more common definition of total differentiability by the existence of a linear approximation.
Concerning the continuity of
$$\Delta_1^{y_0}(x) + \frac{g(x,y) - g(x_0,y)}{x-x_0}$$
at $z_0$, that is unproblematic if $g$ has continuous partial derivatives in a neighbourhood of $z_0$, but if $g$ is merely (totally) differentiable at $z_0$ and behaves badly elsewhere, it need not be the case that that particular choice of $\Delta_k$ works. If for example
$$g(x+iy) = \begin{cases} 0 &, x \neq y^3 \\ y^2 &, x = y^3\end{cases}$$
and $z_0 = 0$, $\Delta_1^{y_0}(x) \equiv 0$, and on the curve $x = y^3$ we have
$$\frac{g(y^3,y) - g(0,y)}{y^3-0} = \frac{1}{y}$$
for $y\neq 0$, which is unbounded in any neighbourhood of $0$.