$\int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} \mathrm{d} x=\ldots$

Given integral: $I = \int_1^3 \frac{\log x^2}{\log (16 x^2 - 8 x^3 + x^4)} \, dx$

Simplify the integrand. The numerator is $\log x^2 = 2 \log x$.
The denominator’s argument factors as $16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(x - 4)^2$, so $\log(x^2(x - 4)^2) = 2 \log x + 2 \log |x - 4|$.
Since $x \in [1, 3]$, $x - 4 < 0$, yielding $|x - 4| = 4 - x$.

Substituting, $I = \int_1^3 \frac{2 \log x}{2 \log x + 2 \log(4 - x)} \, dx = \int_1^3 \frac{\log x}{\log x + \log(4 - x)} \, dx$.

Apply the substitution $x \mapsto 4 - x$ to the integral, giving:
$I = \int_1^3 \frac{\log(4 - x)}{\log(4 - x) + \log x} \, dx$.

Adding these two expressions for $I$:
$2I = \int_1^3 \left( \frac{\log x}{\log x + \log(4 - x)} + \frac{\log(4 - x)}{\log(4 - x) + \log x} \right) dx = \int_1^3 \frac{\log x + \log(4 - x)}{\log x + \log(4 - x)} \, dx = \int_1^3 1 \, dx$.

Thus, $2I = [x]_1^3 = 2$, so $I = 1$.

$\boxed{1}$