How do I prove that $3x^3 +2x + 1 $ is $\omega(x \cdot \log x) $
The easiest way is to check that $\lim_{x \to \infty} \frac{3x^3 + 2x +1}{ x \log x} = +\infty$, which is a sufficient condition for $3x^3 + 2x +1 \in \omega(x \log x)$.

$$
\lim_{x \to \infty} \frac{3x^3 + 2x +1}{ x \log x} =
\lim_{x \to \infty} \frac{3x^3}{ x \log x} = 
\lim_{x \to \infty} \frac{3x^2}{  \log x} =
\lim_{x \to \infty}  6x^2 
= +\infty.
$$