In $\mathbb R$ for a derivative to exist (or a limit generally) it is necessary that the limit be the same in both directions (from below and above) and this is the same in $\mathbb C$ where for a function to be differentiable at a point the difference quotient limit must be the same independent of the way $h$ approach the point, which is where the Cauchy-Riemann equations come from. Why is this only the case in $\mathbb R$ and $\mathbb C$ and not in $\mathbb R^2$? In $\mathbb R^2$ we just make a matrix/vector of partial derivatives and say the derivative is just a linear map. What is the fundamental difference that requires such different ways of defining differentiability?
Real and Complex Analysis – Complex Differentiability vs Differentiability in R2
complex-analysisderivativesreal-analysis
Related Solutions
Well, if a function $f$ is holomorphic (i.e. $\mathbb{C}$-differentiable), it is real analytic, then it has continuous partial derivatives. Therefore, checking the continuity of partial derivatives and then the CR-equations is, in some sense, a perfectly general method in order to determine whether a function is holomorphic or not.
By the way, it is absolutely true that satisfying the CR-equations is not enough: set $f(z)=e^{-z^{-4}}$ for $z\neq0$ and $f(0)=0$. Then $f$ has partial derivatives everywhere, also in the origin, and they satisfy the CR-equations everywhere, but $f$ is not $\mathbb{C}-$differentiable in $0$. Indeed, the partial derivatives are not continuous in $0$.
On a more operative side, there are many other ways to determine if a function is holomorphic or not. For example, if the partial derivatives exist (in the weak sense) and are locally integrable and satisfy the CR-equations, or if it is bounded and holomorphic outside a small enough set. Sums, products, compositions of holomorphic functions are holomorphic, and so on...
Judging from the captions of the pictures, I think we should still talk about real derivatives for a bit.
Brief answer
Neither of functions depicted in your graphs are going to be differentiable at the discontinuities depicted. After you fill in a removable discontinuity of a function like the one on the left, it could be either differentiable or nondifferentiable at the point. Jump discontinuities of functions on the real line are always nondifferentiable, but they might have one-sided derivatives that are well-defined.
Longer anwer
First of all, remember that the derivative at a point is, intuitively, a "limit of slopes as calculated from the left and from the right." From the left you take a limit of $\frac{f(x)-f(x-h)}{h}$ over very small positive values of $h$, and on the right the same happens with $\frac{f(x+h)-f(x)}{h}$. (It can be the case that both of these can be defined, but they don't match and in that case, the derivative isn't defined at that point.)
Notice also that it is critical for $f(x)$ to be defined to carry out these computations, and so you won't get anywhere at all without settling on a value for $f(x)$. If you insist that there's no value for $f(x)$, then the slope is formally undefined. If you are willing to fill in removable discontinuities, though, you can proceed. The derivative may or may not exist after the point is filled in (consider $f(x)=|x|$ with the $x=0$ point removed/replaced.)
That leaves the case of the jump discontinuity, which you've depicted in the right hand picture. Jump discontinuities always make one of the slope limits on the right or on the left jump to infinity. Here's what I mean. Suppose $f(x)$ is anywhere exept filling in the lower circle in your right hand picture. Then as you shrink $h$ in $\frac{f(x+h)-f(x)}{h}$, the associated picture is that of a line which always lies on $(x,f(x))$ and $(x+h,f(x+h)$, which lies on the branch on the right. You see as $h$ shrinks, $x+h$ approaches $x$ from the right. Since $f(x)$ is not on that lower empty circle, this line tips ever more steeply as $h$ shrinks. Thus its slope goes to either $+\infty$ or $-\infty$, and the slope there is undefined.
If $f(x)$ happened to land on that empty lower right circle, then you are guaranteed it wouldn't land on the upper left circle, so you would then deduce that the slope estimate from the left would go off to infinity, and the derivative at the point would again not exist.
Best Answer
A function ${\bf f}:\>{\mathbb R}^2\to{\mathbb R}^2$ is differentiable at $p$ if there is a linear map $A:\>{\mathbb R}^2\to{\mathbb R}^2$ such that $${\bf f}({\bf p}+{\bf h})-{\bf f}({\bf p})=A{\bf h}+o(|{\bf h}|)\qquad({\bf h}\to{\bf 0})\ .\tag{1}$$ This $A$ is then called the derivative of ${\bf f}$ at ${\bf p}$, and is denoted by $d{\bf f}({\bf p})$, or similar. If $d{\bf f}({\bf p})$ exists then the four partial derivatives $f_{i.k}:={\partial f_i\over\partial x_k}({\bf p})$ exist, and the matrix of $d{\bf f}({\bf p})$ with respect to the standard basis in ${\mathbb R}^2$ is the matrix $[f_{i.k}]$.
Let such an ${\bf f}=(f_1,f_2)$ be given. The complexified version $f:=f_1+i f_2$ of this ${\bf f}$ is a differentiable function of the complex variable $z:=x_1+i x_2$ at the "complex point" $p$ iff the matrix of $d{\bf f}({\bf p})$ has the special form $$\left[\matrix{a&-b \cr b& a\cr}\right]\ .$$ The complex number $c:=a+ib$ is then the derivative of $f$ at $p$, and $(1)$ assumes the form $$f(p+h)-f(p)=c\>h+o(|h|)\qquad(h\to0\in{\mathbb C})\ .$$ In short: Complex differentiability is a special case of ${\mathbb R}^2$-differentiability, expressed in terms of other fonts.