Differentiablity of a function $f(z)$ in the complex vs reals I am quite confused about the differentiability of a function in the complex plane vs the real numbers.
Consider the function $f(z) = |z|^2$ for $z \in \mathbb{C}$. Using the definition of a derivative, and taking the limit as $\Delta z \rightarrow 0$, we can see that it is only differentiable at $z = 0$. Elsewhere, the limits are not unique, so do not exist. However, if we consider a similar function $F(x) = |x|^2$ for $x \in \mathbb{R}$, then $F'(x) = 2x$ and exists everywhere in $\mathbb{R}$.
We know that $\mathbb{C}$ is a field extension of $\mathbb{R}$, so $\mathbb{R} \subset \mathbb{C}$. The domain of $f(x)$ is then a subset of the domain of $f(z)$. Why isn't $f(z)$ instead differentiable for all $z = x + i0$?
Likewise for a similar function, $g(z) = \operatorname{Re}(z)$. $g(z)$ is nowhere differentiable in $\mathbb{C}$, yet $G(x) = x$ is differentiable everywhere in $\mathbb{R}$. Why is this the case? Is it due to the limit being used in the definition of a derivative? That the limit requires that $|f(z) - f(z_0)| < \epsilon$, which means that the limit of $f(z)$ must approach $f(z_0)$, as $z \rightarrow z_0$, independent of the direction for the limit to exist. If we were only working with the reals, the limit as $x \rightarrow x_0^+$ must equal that of $x \rightarrow x_0^-$ for the limit to exist, but in the complex plane, the limit must be equal as $z \rightarrow z_0$ from all sides of $z_0$, which is why the derivative for a function in $\mathbb{R}$ can exist, but not a similar one in $\mathbb{C}$?
Does this mean that given two similar functions $f(z)$ and $F(x)$ acting on both $\mathbb{C}$ and $\mathbb{R}$, respectively. Let $f(z=x+iy) = u(x,y) + iv(x,y)$,  $u(x,y) = F(x)$, and $z_0 = x_0 + iy_0$, then
$$f(z) \text{ differentiable at point } z_0 \in \mathbb{C} \text{ and } u'(z) \text{ exists at } z_0 \implies F(x) \text{ differentiable at } 
x_0 \in \mathbb{R}?$$

Why isn't $f(z)$ instead differentiable for all $z = x + i0$?
Because the existence of $\lim\limits_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ is a much stronger requirement than the existence of $\lim\limits_{x \to x_0}\frac{f(x+i0)-f(x_0+i0)}{x-x_0}$. Intuitively, the first limit involves points in "two dimensional neighborhood" of $z_0$ while the second one only involves points lying on a line.
$$f(z) \text{ differentiable at point } z_0 \in \mathbb{C} \text{ and } u'(z) \text{ exists at } z_0 \implies F(x) \text{ differentiable at } 
x_0 \in \mathbb{R}?$$
The answer is yes.
