This integral is not equal to zero.
We may obtain the following closed form.
$$
\begin{align}
\int_0^1 \left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\frac{\log(x)}{x}
\mathrm{d}x & = \dfrac{\ln^2(2\pi)}{4}-\dfrac{\gamma^2}{4}+\dfrac{\pi^2}{48}-\dfrac{\gamma_1}{2}-1\tag1 \\\\
\end{align}
$$
where $\left\{x\right\}$ denotes the fractional part of $x$, $\gamma$ denotes the Euler–Mascheroni constant and where $\gamma_{1}$ denotes the Stieltjes constant defined by
$$
\gamma_{1} = \lim_{N \rightarrow \infty}\left(\sum_{k=1}^{N}\frac{\ln k}{k}-\frac{\ln^{2}N}{2} \right).
$$
Consequently, we have the numerical evaluation:
$$
\begin{align}
\int_0^1 \left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\frac{\log(x)}{x}
\mathrm{d}x = \color{red}{0.00}31782279542924256050500... . \tag2
\end{align}
$$
Here is an approach.
Step 1. Let $s$ be a complex number such that $0<\Re{s}<1$. Then
$$
\int_{0}^{1} x^{s-1}\left\{\frac{1}{x}\right\}\mathrm{d}x = -\frac{1}{1-s} -\frac{\zeta(s)}{s}\tag3
$$ where $\left\{x\right\}$ denotes the fractional part of $x$ and where $\zeta$ denotes the Riemann zeta function.
Proof. Let us assume that $0<\Re{s}<1$. We may write
$$
\begin{align}
\int_{0}^{1} x^{s-1}\left\{\frac{1}{x}\right\}\mathrm{d}x & = \sum_{k=1}^{\infty}
\int_{1/(k+1)}^{1/k} x^{s-1}\left\{\frac{1}{x}\right\}\mathrm{d}x \\
& = \sum_{k=1}^{\infty} \int_{k}^{k+1} \left\{x\right\} \frac{\mathrm{d}x}{x^{s+1}} \\
& = \sum_{k=1}^{\infty} \int_{k}^{k+1} (x-k) \frac{\mathrm{d}x}{x^{s+1}} \\
& = \sum_{k=1}^{\infty} \int_{0}^{1}\frac{v}{(v+k)^{s+1}}\mathrm{d}v \\
& = \sum_{k=1}^{\infty} \int_{0}^{1}\left(\frac{1}{(v+k)^{s}}-\frac{k}{(v+k)^{s+1}}\right)\mathrm{d}v \\
& = \sum_{k=1}^{\infty} \left.\left(\frac{1}{(-s+1)(v+k)^{s-1}} +\frac{k}{s(v+k)^s}\right) \right|_{0}^{1} \\
& = -\frac{1}{1-s}-\frac{\zeta(s)}{s}.
\end{align}
$$
Step 2. We have
$$
\int_{0}^{1} x^{s-1}\left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\log(x)\mathrm{d}x = -\frac{1}{(1-s)^2} +\frac{1}{2s^2} +\frac{\zeta(s)}{s^2} -\frac{\zeta'(s)}{s}. \tag4
$$
Using $(3)$, we readily get
$$
\int_{0}^{1} x^{s-1}\left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\mathrm{d}x = -\frac{1}{1-s}-\frac{1}{2s} -\frac{\zeta(s)}{s}
$$
which we differentiate with respect to $s$ to obtain $(4)$.
Step 3. For $s$ near $0$, we take into account the Taylor series expansion of the Riemann $\zeta$ function:
$$
\begin{align}
& \zeta(s) =-\frac12-\dfrac{\ln(2\pi)}{2} s +\left(\dfrac{\gamma^2}{4}-\dfrac{\pi^2}{48}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{4}+\dfrac{\gamma_1}{2}\right)s^2+\mathcal{O}(s^3)
\\& \zeta'(s) =-\dfrac{\ln(2\pi)}{2} +\left(\dfrac{\gamma^2}{2}-\dfrac{\pi^2}{24}+2\ln(2\pi)-\dfrac{\ln^2(2\pi)}{2}+\gamma_1\right)s+\mathcal{O}(s^2)
\end{align}
$$
and upon letting $s$ tend to $0^+$ in $(4)$ we obtain $(1)$.
Remark: A related result to $(3)$.
\begin{align}J&=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx\\
&=\int_0^1\int_0^1 \frac{x(x^2+1)}{(x^4-x^2+1)(1+xt)}\,dt\,dx\\
&=-\int_0^1\int_0^1 \frac{t(t^2+1)}{(t^4-t^2+1)(1+xt)}\,dt\,dx+\int_0^1\int_0^1 \frac{t^3+t^2x-\sqrt{3}t^2-\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2-\sqrt{3}x+1)}\,dt\,dx+\\
&\int_0^1\int_0^1 \frac{t^3+t^2x+\sqrt{3}t^2+\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2+\sqrt{3}x+1)}\,dt\,dx\\
&=-J+\int_0^1\int_0^1 \frac{t^3+t^2x-\sqrt{3}t^2-\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2-\sqrt{3}x+1)}\,dt\,dx+\\
&\int_0^1\int_0^1 \frac{t^3+t^2x+\sqrt{3}t^2+\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2+\sqrt{3}x+1)}\,dt\,dx\\
\end{align}
Since,
\begin{align}A_1&=\int_0^1 \frac{t}{t^4-t^2+1}\,dt\\
&=\frac{1}{\sqrt{3}}\left[\arctan\left(\frac{2t^2-1}{\sqrt{3}}\right)\right]_0^1\\
&=\frac{\pi}{3\sqrt{3}}\\
A_3&=\int_0^1 \frac{t^3}{t^4-t^2+1}\,dt\\
&=\frac{1}{12}\left[2\sqrt{3}\arctan\left(\frac{2t^2-1}{\sqrt{3}}\right)+3\ln(t^4-t^2+1)\right]_0^1\\
&=\frac{\pi}{6\sqrt{3}}\\
B_1&=\int_0^1 \frac{1}{x^2-\sqrt{3}x+1}\,dx\\
&=2\Big[\arctan\left(2x-\sqrt{3}\right)\Big]_0^1\\
&=\frac{5\pi}{6}\\
B_2&=\int_0^1 \frac{1}{x^2+\sqrt{3}x+1}\,dx\\
&=2\Big[\arctan\left(2x+\sqrt{3}\right)\Big]_0^1\\
&=\frac{\pi}{6}\\
C_1&=\int_0^1 \frac{x}{x^2-\sqrt{3}x+1}\,dx\\
&=\Big[\frac{1}{2}\ln\left(x^2-\sqrt{3}x+1\right)+\sqrt{3}\arctan\left(2x-\sqrt{3}\right)\Big]_0^1\\
&=\frac{1}{2}\ln\left(2-\sqrt{3}\right)+\frac{5}{4\sqrt{3}}\pi\\
C_2&=\int_0^1 \frac{x}{x^2+\sqrt{3}x+1}\,dx\\
&=\Big[\frac{1}{2}\ln\left(x^2+\sqrt{3}x+1\right)-\sqrt{3}\arctan\left(2x+\sqrt{3}\right)\Big]_0^1\\
&=\frac{1}{2}\ln\left(2+\sqrt{3}\right)-\frac{1}{4\sqrt{3}}\pi\\
A_2&=\int_0^1 \frac{t^2}{t^4-t^2+1}\,dt\\
&=\int_0^1 \frac{t^2+1}{t^4-t^2+1}\,dt-\int_0^1 \frac{1}{t^4-t^2+1}\,dt\\
&=\left[\arctan\left(\frac{x}{1-x^2}\right)\right]_0^1-\frac{1}{2\sqrt{3}}\int_0^1 \frac{x}{x^2+\sqrt{3}x+1}\,dx+\frac{1}{2\sqrt{3}}\int_0^1 \frac{x}{x^2-\sqrt{3}x+1}\,dx-\\
&\frac{1}{2}\int_0^1 \frac{1}{x^2+\sqrt{3}x+1}\,dx-\frac{1}{2}\int_0^1 \frac{1}{x^2-\sqrt{3}x+1}\,dx\\
&=\frac{\pi}{2}-\frac{1}{2\sqrt{3}}C_2+\frac{1}{2\sqrt{3}}C_1-\frac{1}{2}B_2-\frac{1}{2}B_1\\
&=\frac{\pi}{4}-\frac{1}{2\sqrt{3}}\ln\left(2+\sqrt{3}\right)\\
A_0&=\int_0^1 \frac{1}{t^4-t^2+1}\,dt\\
&=\int_0^1 \frac{1+t^2}{t^4-t^2+1}\,dt-\int_0^1 \frac{t^2}{t^4-t^2+1}\,dt\\
&=\left[\arctan\left(\frac{x}{1-x^2}\right)\right]_0^1-A_2\\
&=\frac{\pi}{4}+\frac{1}{2\sqrt{3}}\ln\left(2+\sqrt{3}\right)\\
\end{align}
then,
\begin{align}
2J&=\left(\frac{1}{2}A_3B_1+\frac{1}{2}A_2C_1-\frac{\sqrt{3}}{2}A_2B_1-\frac{\sqrt{3}}{2}A_1C_1+\frac{1}{2}A_1B_1+\frac{1}{2}A_0C_1\right)+\\
&\left(\frac{1}{2}A_3B_2+\frac{1}{2}A_2C_2+\frac{\sqrt{3}}{2}A_2B_2+\frac{\sqrt{3}}{2}A_1C_2+\frac{1}{2}A_1B_2+\frac{1}{2}A_0C_2\right)\\
\end{align}
Since,
\begin{align}\frac{1}{2}A_3B_1&=\frac{5\pi^2}{72\sqrt{3}}\\
\frac{1}{2}A_2C_1&=\frac{5\pi^2}{32\sqrt{3}}+\frac{\pi}{16}\ln\left(2-\sqrt{3}\right)-\frac{1}{8\sqrt{3}}\ln\left(2-\sqrt{3}\right)\ln\left(2+\sqrt{3}\right)-\frac{5\pi}{48}\ln\left(2+\sqrt{3}\right)\\
-\frac{\sqrt{3}}{2}A_2B_1&=\frac{5\pi}{24}\ln\left(2+\sqrt{3}\right)-\frac{5\pi^2}{16\sqrt{3}}\\
-\frac{\sqrt{3}}{2}A_1C_1&=-\frac{\pi}{12}\ln\left(2-\sqrt{3}\right)-\frac{5\pi^2}{24\sqrt{3}}\\
\frac{1}{2}A_1B_1&=\frac{5\pi^2}{36\sqrt{3}}\\
\frac{1}{2}A_0C_1&=\frac{\pi}{16}\ln\left(2-\sqrt{3}\right)+\frac{5\pi^2}{32\sqrt{3}}+\frac{1}{8\sqrt{3}}\ln\left(2-\sqrt{3}\right)\ln\left(2+\sqrt{3}\right)+\frac{5\pi}{48}\ln\left(2+\sqrt{3}\right)\\
\frac{1}{2}A_3B_2&=\frac{\pi^2}{72\sqrt{3}}\\
\frac{1}{2}A_2C_2&=\frac{\pi}{12}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{32\sqrt{3}}-\frac{1}{8\sqrt{3}}\ln^2\left(2+\sqrt{3}\right)\\
\frac{\sqrt{3}}{2}A_2B_2&=\frac{\pi^2}{16\sqrt{3}}-\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)\\
\frac{\sqrt{3}}{2}A_1C_2&=\frac{\pi}{12}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{24\sqrt{3}}\\
\frac{1}{2}A_1B_2&=\frac{\pi^2}{36\sqrt{3}}\\
\frac{1}{2}A_0C_2&=\frac{1}{8\sqrt{3}}\ln^2\left(2+\sqrt{3}\right)+\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{32\sqrt{3}}\\
\end{align}
Therefore,
\begin{align}2J&=\frac{\pi}{24}\ln\left(2-\sqrt{3}\right)+\frac{3\pi}{8}\ln\left(2+\sqrt{3}\right)\end{align}
Since,
\begin{align}\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\end{align}
Therefore,
\begin{align}2J&=-\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)+\frac{3\pi}{8}\ln\left(2+\sqrt{3}\right)\\
&=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right)
\end{align}
Thus,
\begin{align}\boxed{J=\frac{\pi}{6}\ln\left(2+\sqrt{3}\right)}
\end{align}
Best Answer
Since the integrand is an even function, we can write
$$ I = \frac{1}{2}\int_{-1}^{1} \log\left(\frac{1-x}{1+x}\right)\log\left(\frac{1-x^2}{1+x^2}\right)\,\frac{dx}{x}. $$
Now deforming the line contour $[-1, 1]$ to the semicircular contour from $-1$ to $1$ and substituting $x = e^{i\theta}$,
$$ I = -\frac{i}{2} \int_{0}^{\pi} \log(-i\tan(\theta/2)) \log(-i\tan \theta) \, d\theta, $$
where we utilized the identity $\frac{1-e^{i\theta}}{1+e^{i\theta}} = -i\tan(\theta/2)$. Now we note that, for $\theta \in (0, \pi/2) \cup (\pi/2, \pi)$,
$\log(-i\tan(\theta/2)) = \log\tan(\theta/2) - \frac{i\pi}{2}$,
$\log(-i\tan\theta) = \log\lvert\tan\theta\rvert - \operatorname{sign}(\tan\theta)\frac{i\pi}{2}$.
$\int_{0}^{\pi} \log\lvert\tan\theta\rvert \, d\theta = 2 \int_{0}^{\pi/2} (\log\sin\theta - \log\cos\theta) \, d\theta = 0$.
Plugging these back and taking real parts only (since we know that $I$ is real),
\begin{align*} I &= -\frac{\pi}{4} \int_{0}^{\pi} \left( \log\lvert\tan\theta\rvert + \operatorname{sign}(\tan\theta)\log\tan(\theta/2) \right) \, d\theta \\ &= -\frac{\pi}{2} \int_{0}^{\pi/2} \log\tan(\theta/2) \, d\theta \\ &= -\pi \int_{0}^{1} \frac{\log u}{1+u^2} \, du, \qquad (u=\tan(\theta/2)) \\ &= \pi C. \end{align*}
Generalization. Utilizing a similar idea s in the computation above, we can prove that
Of course, the above can be simplified further by using either Hurwitz zeta function or trigamma function .