Note that $1-x^4=(1-x)(1+x+x^2+x^3)$. Thus, we have
$$\begin{align}
\int_0^\infty \log(x) \frac{1-x}{1-x^4}\,dx&=\int_0^\infty \frac{\log(x)}{(1+x)(1+x^2)}\,dx\\\\
\end{align}$$
RESIDUES:
Now, let's evaluate the integral of $\frac{\log^2(z)}{(1+z)(1+z^2)}$ around the classical keyhole contour (shown in the diagram of the posted question). Proceeding, the residue theorem guarantees that
$$\oint_{C_{kh}}\frac{\log^2(z)}{(1+z)(1+z^2)}\,dz=2\pi i \sum_{i=1}^3 \text{Res}\left(\frac{\log^2(z)}{(1+z)(1+z^2)},z=z_i\right)$$
where $z_1=-1$, $z_2=i$, and $z_3=-i$. Evaluating the residues, reveals
$$\begin{align}\text{Res}\left(\frac{\log^2(z)}{(1+z)(1+z^2)},z=-1\right)&=\frac{-\pi^2}{2}\tag{1a}\\\\
\text{Res}\left(\frac{\log^2(z)}{(1+z)(1+z^2)},z=i\right)&=\frac{-\pi^2/4}{(1+i)2i}\tag{1b}\\\\
\text{Res}\left(\frac{\log^2(z)}{(1+z)(1+z^2)},z=-i\right)&=\frac{-9\pi^2/4}{(1-i)(-2i)}\tag{1c}
\end{align}$$
CONTOUR INTEGRAL:
In addition, we see that
$$\begin{align}
\oint_{C_{kh}}\frac{\log^2(z)}{(1+z)(1+z^2)}\,dz&=\int_0^\infty \left(\frac{\log^2(x)}{(1+x)(1+x^2)}-\frac{(\log(x)+i2\pi)^2}{(1+x)(1+x^2)}\right)\,dx\\\\
&=-i4\pi \int_0^\infty \frac{\log(x)}{(1+x)(1+x^2)}\,dx\\\\&+4\pi^2\int_0^\infty \frac{1}{(1+x)(1+x^2)}\,dx\tag2
\end{align}$$
Note that it is a straightforward exercise (left to the reader) to show that the contributions to the contour integral from integration around the semi-circles at $|z|=R$ and $|z|=\varepsilon$ vanish as $R\to \infty$ and $\varepsilon \to 0^+$, respectively.
Putting together $(1a)-(1c)$ and $(2)$, we find that
$$\int_0^\infty \frac{\log(x)}{(1+x)(1+x^2)}\,dx=-\pi^2/16$$
Aside:
As a bonus, we also find that
$$\int_0^\infty \frac{1}{1+x+x^2+x^3}\,dx=\pi/4 $$
