Given are the complex number $z = x + iy$ (with $z \in \mathbb{C}$, $i^2 = -1$ and $x, y \in \mathbb{R}$), and the function $$f(z) = f(x +iy) = x^3+y^2+i(-6x+2y)$$
I've heard that this function is is only differentiable for the complex numbers $z_{1,2} = \pm \sqrt{2\over{3}} - 3i$. In other words, $f'(z) = \frac{df(z)}{dz}$ is defined only for $z = \pm \sqrt{2\over{3}} - 3$.
(In this question I'm not worried about the derivation.)
My question is, what is the intuition behind this? If you have a function $f(x) = x^2$ then $f'(x) = 2x$ is the rate of change of $f$. Or if a car drives with a velocity of $v(t)$, the car has an acceleration of $a(t) = v'(t)$.
A function $f(x) = |x|$ cannot be differentiated at $x = 0$ because the rate of change of $f$ changes at once from $-1$ to $1$ if we cross $x=0$ when coming from $x < 0$ to $x > 0$.
But how can a function have a continuous "change of speed" at only two points, and not at all other points. Or how else should I interpret this?
Keep in mind that being differentiable as a complex function (ie holomorphic) is a stronger condition than being differentiable as a real function; it must preserve angles. Your function $f$ can be identified with the function $g$ on the real plane given by $$ g(x,y)=(x^3+y^3,-6x+2y). $$ Note that $g$ is differentiable everywhere, but if you imagine applying $g$ to a grid in the plane, there is no reason to expect the grid lines to stay perpendicular everywhere.
Indeed we can describe the derivative of $g$ as a $2\times2$ real matrix. To be holomorphic at a point, this matrix must be a rotation-dilation. Since the space of those is $2$ dimensional and the space of $2\times2$ matrices is $4$ dimensional, this intuitively imposes $2$ conditions, so for a generic real function we'd expect it to happen at a discrete set of points.