Why isn't $f(z)=\bar{z}$ complex differentiable

Viewed as a real map the complex conjugation map $z \mapsto \bar z$ is differentiable (in fact, linear): Under the usual identification $\Bbb R^2 \leftrightarrow \Bbb C$, $(x, y) \leftrightarrow x + iy$, it is given by $(x, y) \to (x, -y)$.

On the other hand, it is not complex-differentiable, which is a much stronger property. By definition, $f$ is complex-differentiable at $z_0$ if the usual limit $$\lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$$ of the difference quotient of $f$ exists, in which case we define $f'(z_0)$ to be its value. Critically, $h$ is here a complex variable: In the case of the conjugation map, the limit simplifies to $$\lim_{h \to 0} \frac{\bar h}{h},$$ and for real $h$ we have $\frac{\bar h}{h} = \frac{h}{h} = 1$, but for imaginary $h$ we have $\frac{\bar h}{h} = \frac{-h}{h} = -1$. Thus, the limit does not exist for any $z_0$, and the map is not complex-differentiable anywhere.

Of course it doesn't satisfy one of the possible definitions (Cauchy-Riemann equations, the specific limit does not exist), but in my opinion that doesn't give you much intuition - at least of the geometric kind. Hopefully the following will compensate for that.

TL;DR:

If $f$ is differentiable at a point $z_0$, then there's the linear function $$(Df)(z_0)\colon \mathbb C \to \mathbb C;\quad z \mapsto f'(z_0)z $$ that approximates $f$ well around $z_0$. Multiplication by a complex number is a rotation or a scaling of the complex plane, thus it keeps orientation.

These imply that $f$ has to keep orientation locally, around $z_0$. Conjugation is a reflection so it flips orientation, therefore it cannot be differentiable at any point in the complex sense.

If $f\colon \mathbb C \to \mathbb C$ is complex-differentiable, then it is differentiable as an $\mathbb R^2 \to \mathbb R^2$ function, and it locally almost keeps angles and circles. To see this, look at one equivalent version of the complex-differentiability: $$f(z_0 + re^{it}) = f(z_0) + re^{it}f'(z_0) + \omicron(r), $$ where $z_0 \in \mathbb C, 0 \leq r, 0 \leq t \leq 2\pi$. In fact, as a demonstration have a picture of the image of two circles around $1 + i$ by $\operatorname{id_{\mathbb C}}^2$.

The black,$\{z_0 + re^{it} \mid t \in [0,2\pi]\}$ ones are our original circles ,
the reds, $\,\,\color{red}{f(\{z_0 + re^{it} \mid t \in [0,2\pi]\})}$ are the blacks' images by $f$, and
the blues, $\color{blue}{\{f(z_0) + re^{it}f'(z_0) \mid t \in [0,2\pi] \}}$ are the first order approximation of the red ones.

If we pick $z_0 = 1 + i, r=0.1$ and $0.2$, then we get, in order:

As you can see in the $r=0.1$ case the red and blue circles are almost equal. As $r \to 0$, the difference becomes smaller and smaller. That's what I meant by locally almost keeping circles.

Here's another picture, displaying the image of a point from each of the two black circles. The black and blue circles represent the same as before, the green circles are due to the $\omicron(r)$ part.

Look for the points $\color{red}{a}, \color{blue}{a'}, \color{red}{f(a)}$ and $\color{purple}{b},\color{blue}{b'}, \color{purple}{f(b)}$. In fact, in the $r=0.1$ case the green circle is so small, you can barely see any of it (this is why the red and blue circles from a moment ago were almost equal).

As you can see, the image of a circle is essentially the sum of multiple circles, in this case, two; however, if $f$ has lots of non-zero derivatives, then there'll be lots of circles, and each "new" circle will have smaller radius, but higher frequency than the previous one (assuming $r < 1$).

If you know about Fourier-series, then this concept might sound familiar. By the way, if you haven't already, read this answer.

Anyway, this is not so surprising: if you consider the Taylor series of a holomorphic $f$ at a point, and plug in a point from a circle around that point, you'll get $$ f(z_0 + re^{it}) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}r^ne^{int}.$$ Here the $n$th term will correspond to the circle $\left\{\frac{f^{(n)}(z_0)}{n!}r^ne^{int} \mid t \in [0,2\pi]\right\}$. OK, back to the original question.

Hopefully I've convinced you that the $\omicron(r)$ part is not significant. (Of course only so if the first derivative is non-zero, but in the other cases we can say something similar too.)

Let's say a word about locally almost keeping angles. Remember $$f(z_0 + re^{it}) = f(z_0) + re^{it}f'(z_0) + \omicron(r).$$ Picking two numbers, $z_0 + re^{it}, z_0 + re^{is}$ the angle between them that is kept locally is $(t-s)$. Their images are $$ f(z_0) + re^{it}f'(z_0) + \omicron(r), \text{and } f(z_0) + re^{is}f'(z_0) + \omicron(r). $$ As you can see, the local (i.e. looking from the point $f(z_0)$) angle between them is again, approximately $(t-s)$.

Now let's really get back to your original question. For $f(z) = \bar z$, assuming differentiability at $z_0$, the following would be true for all $t\in [0,2\pi]$: \begin{align} f(z_0 + re^{it}) &= f(z_0) + re^{it}f'(z_0) + \omicron(r) \\ \iff re^{-it} &= re^{it}f'(z_0) + \omicron(r). \end{align} This is absurd. Of course we could also note that conjugation locally flips angles. Going back to our points $z_0 + re^{it}, z_0 + re^{is}$, the angle is still $(t-s)$. Their images now are $$\bar{z_0} + re^{-it}, \text{ and } \bar{z_0} + re^{-is}, $$ meaning that the appropriate angle is $-(t-s)$. Thus again, conjugation can not be complex-differentiable.

$f$ does not create any discontinuities ! $f$ is continuous, but not complex differentiable.

Let $z_0 \in \mathbb C$ and $t \in \mathbb R$. Then:

$\frac{f(z_0+t)-f(z_0)}{t}=1 \to 1 $ for $t \to 0$,

but

$\frac{f(z_0+it)-f(z_0)}{it}=-1 \to -1 $ for $t \to 0$.