Derivative of $\ln (z), z\in\mathbb{C}$ Let $f(z) = \ln z := \ln |z| + \arg (z)i$. Then the derivative is (if it exists) by definition:
$$\lim_{h\to 0}\frac{\ln (z+h)-\ln (z)}{h}=\lim_{h\to 0}\frac{\ln |z+h| +\arg(z+h)i-\ln |z| -\arg(z)i }{h}$$
 A: Let $ U =\mathbb C \setminus (-\infty,0].$ Then $f(z)=\ln |z| + i\arg z$ is continuous on $U$ and we have $e^{f(z)} = z$ there. This shows $f(z)$ is injective on $U.$ Fix $z\in U.$ Then for small nonzero $h$ we have
$$1=\frac{e^{f(z+h)}-e^{f(z)}}{h} = \frac{e^{f(z+h)}-e^{f(z)}}{f(z+h) - f(z)}\frac{f(z+h) - f(z)}{h}.$$
The injectivity of $f$ shows that $f(z+h) - f(z)\ne 0,$ so we're OK dividing by it above. As $h\to 0, f(z+h) \to f(z)$ by the continuity of $f.$ So the first difference quotient on the right tends to the derivative of $e^w$ at $w=e^{f(z)},$ which is $e^{f(z)} = z \ne 0.$ Knowing all of this, we can now write
$$\tag 1 \frac{f(z+h) - f(z)}{e^{f(z+h)}-e^{f(z)}} = \frac{f(z+h) - f(z)}{h}.$$
Since the left side of $(1) \to 1/z,$ we get $f'(z) = 1/z$ (as expected).
A: I'll use $\log$ for the complex logarithm and $\ln$ for the real-valued logarithm; you then have:
$$\log z = \ln |z| + i \arg z = \ln r + i\varphi$$
where I use $r = |z|$ and $\varphi = \arg z$ for simplicity. In this form, we have written
$$\log z = u(x,y)+iv(x,y)$$
with  $u(x,y) = \ln r = \ln \sqrt{x^2+y^2} = \tfrac{1}{2}\ln(x^2+y^2)$ and $v(x,y) = \varphi$.
You can check the Cauchy-Riemann equations yourself and find (*)
$$\frac{\partial u}{\partial x} = \frac{x}{x^2+y^2} = \frac{\partial \varphi}{\partial y} \quad ; \quad \frac{\partial u}{\partial y} = \frac{y}{x^2+y^2} = - \frac{\partial \varphi}{\partial x}$$
Use this to find the derivative directly:
$$\frac{\mbox{d} \log z}{\mbox{d}z} = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} = \frac{x-iy}{x^2+y^2} = \frac{1}{z}$$
Remark: depending on how you define the complex logarithm, there will be different ways to find its derivative.

(*) The derivatives for $u$ are easy. For $v$, we have:
$$\left\{ \begin{array}{ccc}
x = r\cos\varphi \\
y = r\sin\varphi
\end{array}\right.$$
where both $r$ and $\varphi$ are (implicit) functions of $x$ and $y$. Differentiating both equations w.r.t. $x$ gives:
$$\left\{ \begin{array}{ccc}
1 = \cos\varphi\frac{\partial r}{\partial x}-r\sin\varphi\frac{\partial \varphi}{\partial x} \\
0 = \sin\varphi\frac{\partial r}{\partial x}+r\cos\varphi\frac{\partial \varphi}{\partial x}
\end{array}\right.$$
This is a linear system of two equations in the variables $\frac{\partial r}{\partial x}$ and $\frac{\partial \varphi}{\partial x}$; solve for $\frac{\partial \varphi}{\partial x}$. In the same way, take the derivative w.r.t. $y$ and solve for $\frac{\partial \varphi}{\partial y}$. Can you take it from here?
