One of three of a kind - Part (1)

$\displaystyle \sum_{k=1}^{\infty}{\frac{\phi(k)}{k} \ln(1-x^k)} = - \sum_{k=1}^{\infty}{\frac{\phi(k)}{k} (-\ln(1-x^k))}$

$\displaystyle - \sum_{k=1}^{\infty}{\frac{\phi(k)}{k} \sum_{m=1}^{\infty}{\frac{x^{km}}{m}}}$

Now, you can just use Dirichlet series by utilizing Dirichlet convolution. But we'll do this in a beginner's way. We'll write our series in the following way:

$\displaystyle - \sum_{n=1}^{\infty}{\sum_{a|n}{\frac{\phi(a)}{am}} x^n}$

where $am = n$ so it simply becomes

$\displaystyle - \sum_{n=1}^{\infty}{\sum_{a|n}{\phi(a)} \frac{x^n}{n}}$

Now, using Gauss' argument that $\displaystyle \sum_{a|n}{\phi(a)} = n$ [will not prove here]

$\displaystyle = - \sum_{n=1}^{\infty}{x^n} = \frac{-x}{1-x}$

Let $\large f(x)=\sum_{k=1}^\infty \frac{\phi(k)}k \ln(1 - x^k)$ This implies $\large f'(x)=\sum_{k=1}^\infty \frac{\phi(k)}{k}\frac{-kx^{k-1}}{ 1 - x^k}=-\frac{1}{x} \sum_{k=1}^\infty \frac{\phi(k)x^{k}}{1 - x^k}$ Using Lambert series which states for $|x|<1$ $\large \sum_{k=1}^\infty \frac{\phi(k)x^{k}}{1 - x^k}= \frac{x}{(1-x)^2}$ Thus $\large f'(x)=-\frac{1}{(1-x)^2}\implies f(x)=-\frac{1}{1-x}+C$ We evaluate $C$ by plugging in $x=0$ to get $C=1$ and finally $\large f(x)=-\frac{1}{1-x}+1=\boxed{\frac{-x}{1-x}}$