Let $u=u(x)$ be a real-valued function defined on $\mathbb R$.
How does this inequality hold?
$$ \| u^3 \|_{H^3} \leq C ( \| u \|_{L^\infty}^2 + \| \partial_x u \|_{L^\infty}^2 ) \| u \|_{H^3}$$
Here $H^s$ denotes the usual Sobolev space, $\| \cdot \|_{H^s} = \| \cdot \|_{H^s (\mathbb R)}$
The nice thing about working on $\mathbb R$ is that all Sobolev functions have locally absolutely continuous representatives, of which we take derivatives in the usual way, and they exist at almost every point.
So, $u$ is a bounded function with locally absolutely continuous second derivative $u''$, and such that $u'''$ (which exists a.e.) is square integrable. The usual calculus rules apply to $u^3$:
$$(u^3)'' = (3u^2 u')' =(3u^2)'u'+(3u^2)u''= \dots $$ and I guess I'm going to let you finish this. The point is, $(u^3)''$ is locally absolutely continuous.
Similarly, $(u^3)'''$ exists at every point where $u'''$ exists, and is calculated in the usual way. The integral inequality is a consequence of the triangle inequality and the usual estimate $$\int |fg|\le \|f\|_{L^\infty}\int|g|$$