Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote 
$y = -2x - 3/2$.
How we get this asymptote?

You can apply simple algebraic transformations, being careful to always have positive expressions under the square root. For $x<0$, and thus $|x|=-x$, the expression is equal to
\begin{align}
=\frac{-x\sqrt{-x}}{\sqrt{3-x}}-x
&=\frac{x}{\sqrt{3-x}}(-\sqrt{-x}-\sqrt{3-x})\\
\\
&=-2x+\frac{x}{\sqrt{3-x}}(-\sqrt{-x}+\sqrt{3-x})
\\
&=-2x-\frac{-x}{\sqrt{3-x}}\frac{3}{\sqrt{-x}+\sqrt{3-x}}
=-2x-\frac{3}{\sqrt{1-\frac3x}+1-\frac3x}
\end{align}
which asymptotically is equal to $-2x-\frac32$ for $x\to-\infty$.

For $x>3$ the expression is equal to
$$
\frac{x}{\sqrt{x-3}}(\sqrt{x}-\sqrt{x-3})
=\frac{x}{\sqrt{x-3}}\frac{x-(x-3)}{\sqrt{x}+\sqrt{x-3}}
=\frac{3}{\sqrt{1-\frac3x}+1-\frac3x}
$$
which asymptotically converges to the constant $\frac32$.

