Consider:$$f:\mathbb{R}^2\supset X\rightarrow\mathbb{R}^2,~z=(x,y)\mapsto \zeta=(\xi, \eta)=f(x,y)$$
Let it be $\mathbb{R}$-differentiable.
If you displace a point $z$ from a $z_0$ the linear part of the change in the value of $f$ (what is called the differential of $f$ at $z_0$) goes from zero vector to a vector that is $\mathbb{R}$-linearly related to $z-z_0$, that is $$df_{z_0}(\Delta z) = D_{z_0}\Delta z\tag{1}$$ where $\Delta z=z-z_0$, $D_{z_0}$ is a $\mathbb{R}$-linear operator.
(You can skip this but remember that the differential of $f$ is defined in $\mathbb{R}^2$ and not simply in $X$ as is $f$. The differential is a linear map and among other things its domain must be a linear space and not a subset. Indeed its independent variable usually indicated with $dz$ is a displacement of the independent variable of $f$ and it can be taken greater as you want even if the resultant displaced $z$ goes outside of $X$: that's why some highlights this fact by calling it virtual displacement. When the virtual displacement is such that $z$ remains in $X$ then the real diplacement $\Delta z$ exists and $\Delta z=dz$)
$\mathbb{R}$-conclusion.
You can easily see in the $\zeta$-plane where $df_{z_0}(\Delta z)$ is sent when $\Delta z$ is $\mathbb{R}$-linearly related to $\Delta z'$ and $\Delta z''$, that is $\Delta z=a\Delta z'+b\Delta z''$ with $a,~b\in\mathbb{R}$, knowing $df_{z_0}(\Delta z')$ and $df_{z_0}(\Delta z'')$ for every $\Delta z'$ and $\Delta z''$, because for $\mathbb{R}$-linearity it must be that $df_{z_0}(a\Delta z' +b\Delta z'') = a\cdot df_{z_0}(\Delta z')+b\cdot df_{z_0}(\Delta z'')$.
Now let $f$ be $\mathbb{C}$-differentiable.
What I've said above before "$\mathbb{R}$-conclusion" remains unaltered with the exception of the occurrences of the expression $\mathbb{R}$-linear (included all grammatical variations) that must be substituted for by $\mathbb{C}$-linear.
Now from $(1)$ above, due to the fact each $\mathbb{C}$-linear operator on $\mathbb{C}$-monodimensional spaces is characterized completely by the multiplication by a complex scalar and so by two real parameters: a relative length variation and angular displacement applied to each and every vector, it follows that the vector $df_{z_0}(\Delta z)$ has a relative length variation and angular displacement with respect to $\Delta z$ that are independent from $\Delta z$! (while this was not the case when only $\mathbb{R}$-differentiability was involved).
$\mathbb{C}$-conclusion.
You can easily see in the $\zeta$-plane where $df_{z_0}(\Delta z)$ is sent when $\Delta z$ is $\mathbb{C}$-linearly related to $\Delta z'$, that is $\Delta z = a\Delta z'$ with $a\in\mathbb{C}$, knowing $df_{z_0}(\Delta z')$ for every $\Delta z'$, because for $\mathbb{C}$-linearity it must be that $df_{z_0}(a\Delta z') = a\cdot df_{z_0}(\Delta z')$.
(That is, if $\Delta z$ has a length $k$ times that of $\Delta z'$ and an angle of $\theta$ w.r.t. $\Delta z'$ then $df_{z_0}(\Delta z)$ will have a length $k$ times that of $df_{z_0}(\Delta z')$ and an angle of $\theta$ w.r.t. $df_{z_0}(\Delta z')$. That's the meaning of $\mathbb{C}$-linearity in $\mathbb{C}$-dimension $1$, that is, $\mathbb{C}$-proportionality)
But in general you cannot see using the same strategy where $df_{z_0}(\Delta z)$ is sent knowing $df_{z_0}(\Delta z')$ and knowing that $\Delta z$ and $\Delta z'$ are only $\mathbb{R}$-linearly related, that is $\Delta z = h(\Delta z')$, where $h$ is an $\mathbb{R}$-linear operator on $\mathbb{C}$. Indeed, this operator is completely determined when it is known what values it takes on two $\mathbb{R}$-linearly independent vectors. Let's say that it is know, for instance, what the values taken on the vectors $1$ and $i$ are (but we can choose other values). Now let's $\mathbb{R}$-decompose $\Delta z'$ w.r.t. $1$ and $i$: $\Delta z'=\frac{\Delta z'+\overline{\Delta z'}}{2}+\frac{\Delta z'-\overline{\Delta z'}}{2} =\frac{\Delta z'+\overline{\Delta z'}}{2}+i\frac{\Delta z'-\overline{\Delta z'}}{2i}$. For this to be a $\mathbb{R}$-decomposition w.r.t. to $1$ and $i$, the coefficients of $1$ and $i$ must be real, and they are if $\overline{\Delta z'}$ is the complex conjugate of $\Delta z'$. Then it follows straighforwardly that $$h(\Delta z') = h(1)\frac{\Delta z'+\overline{\Delta z'}}{2} + h(i)\frac{\Delta z'-\overline{\Delta z'}}{2i} = \frac{h(1)-i\cdot h(i)}{2} \Delta z' + \frac{h(1)+i\cdot h(i)}{2}\overline{\Delta z'}$$
Concluding, it is:
$$df_{z_0}(\Delta z) = df_{z_0}(h(\Delta z')) = \frac{h(1)-i\cdot h(i)}{2} df_{z_0}(\Delta z') + \frac{h(1)+i\cdot h(i)}{2}df_{z_0}(\overline{\Delta z'})$$ that means that knowing a given $\mathbb{R}$-linear relation between $\Delta z$ and $\Delta z'$ you cannot determine the value of $df_{z_0}(\Delta z)$ from just $df_{z_0}(\Delta z')$: you need $df_{z_0}(\overline{\Delta z')}$ as well when $\frac{h(1)+i\cdot h(i)}{2}\ne0$, so in this case $df_{z_0}$ is not $\mathbb{R}$-linear (although it is $\mathbb{C}$-linear).
End of the $\mathbb{C}$-conclusion.
As you have seen, differences between $\mathbb{R}$- and $\mathbb{C}$-differentiability are to be searched for in the algebraic part of the process not the analytical and recognizing the special character of the $\mathbb{C}$-differentiability from the fact that composition of a $\mathbb{C}$-linear operator (the differential) with a $\mathbb{R}$-linear operator is only a $\mathbb{R}$-linear operator.
