$$f(n) = n^2 \log n, \qquad g(n) = n \log^{10} n$$
Dividing both these functions by $(n \log n)$, we get:
$$F(n) = n, \qquad G(n) = \log^9 n$$
Since any polynomial function grows faster than any logarithmic function, we have:
$F(n)$ grows faster than $G(n)$
$\implies$ $f(n)$ grows faster than $g(n)$
$\implies n \log^{10} n = O(n^2 \log n)$