Is the complex function $f(z) = Re(z)$ differentiable?

The definition of derivative can be written as

$$ f'(z) = \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} $$

which looks just like the real-variable definition, but here this is taken in the complex sense, i.e. $h$ is allowed to be a complex number. $h \to 0$ means the distance in the complex plane from $h$ to $0$ goes to $0$ (or equivalently, both real and imaginary parts of $h$ go to $0$). In order for the limit to exist, you must always get the same value as $h \to 0$ in any manner.

If you take $h$ to be real, $f(z+h) = f(z) + h$ and the quotient is $1$.

If you take $h$ to be imaginary, $f(z+h) = f(z)$ and the quotient is $0$.

The limit as $h \to 0$ doesn't exist: it can't be both $1$ and $0$. Thus we say $f'(z)$ doesn't exist, and the function is not differentiable.

For a function to be differentiable in $\Bbb{C}$, it must satisfy the Cauchy-Riemann equations, that is, if $$f(x,y)=u(x,y)+iv(x,y)$$ it must satisfy $$u_x=v_y\\u_y=-v_x$$

But for $f(z)=\Re(z)=x$ we get $$u_x=1\neq v_y=0$$

So it is not differentiable.

Another way to see it, it is that the real part of a complex number can be written with its conjugate: $Re(x) = \frac{1}{2} (x + x^*)$. Since the conjugate function is the classical example of a non-complex-differentiable function (see for exampe this), it follows that the real part is not complex-differentiable.