Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$
Well,
$$\sqrt{x^6 - 9 x^3} = x^{3} \left(1 - \frac{9}{x^{3}}\right)^{1/2}$$
and so the question is really about understanding how $\sqrt{1 - t}$ looks when $t$ is pretty small. By noticing that the derivative $\frac{d}{dt} \sqrt{t} = \frac 1 2$ when $t = 1$, we can say that
$$\left(1 - \frac{9}{x^{3/2}}\right)^{1/2} \approx \frac 1 2 \left(\frac{-9} {x^{3}}\right)$$
which leads quickly to the claimed limit of $-9/2$.
Alternatively, there is a completely standard technique of multiplying and dividing by the conjugate, which is $\sqrt{x^6 - 9x^3} + x^3$, but I find that this isn't very enlightening.
$$\sqrt{x^6-9x^3}-x^3=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3}$$
As $x$ goes to $\infty$, $\sqrt{x^6-9x^3} \approx x^3.$
More, precisely, divide the numerator and denominator by $x^3$:
\begin{align} \sqrt{x^6-9x^3}-x^3&=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}\\&=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3} \\ & =\frac{-9}{\sqrt{1-9x^{-3}}+1} \end{align}