A differential equation $y^{\prime \prime}+\dfrac{y^{\prime}}{z}=0$ with the initial conditions $y(1)=0$ and $y^{\prime}(1)=1$. Power series approach will lead writing $y = f(z)$ as a series centered at $z = 1$. The solution (given) is the well-known complex-logarithm series $$f(z):=\sum_{j=1}^{\infty} \frac{(-1)^{j-1}(z-1)^{j}}{j}$$
My Approach : The DE can be re-written as $f''(z)= - 1/z f'(z)$ with $f(1) = 0$ and $f'(1)=1$, it is a matter of finding the subsequent derivatives evaluated at $z=1$: $f''(1) = -1$ and using the quotient rule $$f'''(z) = \dfrac{-z f''(z) + f'(z)}{z^2} = -\dfrac{f''(z)}{z} + \dfrac{f'(z)}{z^2} \implies f'''(1) = -2$$ Continuing for the fourth derivative yields $$f^{(4)}(z) = -\dfrac{2}{z^3} f'(z) + \dfrac{2}{z^2} f''(z) - \dfrac{1}{z} f'''(z)$$
From these I don't see how to get the general expression $f^{(j)}(z) = (-1)^{(j-1)}/j$. Any help in resolving this issue is much appreciated.
It may help to write $z = t+1$, so you're looking for a series in powers of $t$. Moreover, since the equation only depends on $f'$ and $f''$, not $f$, write $f'(z) = g(t)$; you can integrate term-by-term at the end. Now write the equation as $$ 0 = z f''(z) + f'(z) = (t+1) g'(t) + g(t) $$ with $$g(t) = \sum_{j=0}^\infty c_j t^j,\ c_0 = f'(1) = 1$$ Next, what is the coefficient of $t^j$ in $(t+1) g'(t) + g(t)$?