Prove $f(z)=-\frac{\log(1-z)}{z}$ is convex in $\mathbb{D}$

Here is a proof which makes use of Maple, namely, its DirectSearch package created by Sergey Moiseev. The GlobalOptima command finds the global minimum of $${\Re}\left(1+z\dfrac{f''(z)}{f'(z)}\right)$$ in the unit disk with the absolute error less than or equal to $10^{-12}$:

f := -log(1-z)/z:
A := evalc(Re(eval(1+z*(diff(f, z, z))/(diff(f, z)), z = x+I*y))):
DirectSearch:-GlobalOptima(A, {x^2+y^2 <= 1}, tolerances = 10^(-12));

[ 0.294349724781047,[x=- 0.999999999999916,y= 0.0000004101682696],498]

B := evalc(Re(eval(1+z*(diff(f, z, z))/(diff(f, z)), z = x+I*sqrt(-x^2+1)))):
DirectSearch:-GlobalOptima(B, {x >= -1, x <= 1}, tolerances = 10^(-12));

[ 0.294349724781044,[x=- 0.999999999999999],126]

These outputs mean that the global minimum equals $0.2943497247810$. I am pretty sure that the results imply the affirmative answer to the question. However, colleagues are welcome to discuss it.

Addition. Just for illustration, here is a plot of A on the unit disk.

plot3d(A, y = -sqrt(-x^2+1) .. sqrt(-x^2+1), x = -1 .. 1, axes = frame);

In fact, the function is unbounded from above around $z=1$.