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?

Let
$$
f(x)
=\sqrt{\frac{x^3}{x-3}}-x
=|x|\sqrt{\frac{x}{x-3}}-x
$$
For $x<0$ we have $|x|=-x$, thus
$$
f(x)
=-x\sqrt{\frac{x}{x-3}}-x
=\Bigl(-\sqrt{\frac{x}{x-3}}-1\Bigr)x
=\Bigl(-\sqrt{\frac{1}{1-3/x}}-1\Bigr)x.
$$
For $x<0$;
$$\lim_{x\to-\infty}\frac{f(x)}{x}
=\lim_{x\to-\infty}\Bigl(-\sqrt{\frac{1}{1-3/x}}-1\Bigr)
=-1-1=-2.
$$
Now, study
\begin{align*}
f(x)-(-2x)
&
=f(x)+2x
=-x\sqrt{\frac{x}{x-3}}-x+2x
=-x\sqrt{\frac{x}{x-3}}+x
=\Bigl(-\sqrt{\frac{x}{x-3}}+1\Bigr)x.
\end{align*}
Maclaurin expansion gives
\begin{align*}
\sqrt{\frac{x}{x-3}}
&=\sqrt{\frac{x-3+3}{x-3}}
=\sqrt{1+\frac{3}{x-3}}
=1+\frac{1}{2}\cdot\frac{3}{x-3}+\Bigl(\frac{3}{x-3}\Bigr)^2L(1/x)
\end{align*}
and we have
\begin{align*}
\Bigl(-\sqrt{\frac{x}{x-3}}+1\Bigr)x
&=\Bigl(-\Bigl(1+\frac{1}{2}\cdot\frac{3}{x-3}+\Bigl(\frac{3}{x-3}\Bigr)^2L_1(1/x)\Bigr)+1\Bigr)x
\\&=\Bigl(-1-\frac{1}{2}\cdot\frac{3}{x-3}-\frac{1}{(x-3)^2}L_2(1/x)+1\Bigr)x
\\&=\Bigl(-\frac{3}{2}\cdot\frac{1}{x-3}-\frac{1}{(x-3)^2}L_2(1/x)\Bigr)x
\\&=-\frac{3}{2}\cdot\frac{x}{x-3}-\frac{x}{(x-3)^2}L_2(1/x)
\\&=-\frac{3}{2}\cdot\frac{1}{1-3/x}-\frac{x}{(x-3)^2}L_2(1/x)
\\&\to-\frac{3}{2}
\end{align*}
as $x\to-\infty$, where $L_{1,2}(1/x)$ is limited for large $x$.
Hence
$$
\lim_{x\to-\infty}\bigl(f(x)+2x\bigr)=-\frac{3}{2}
$$
showing that
$$
f(x)\,\,\mathrm{"}=\mathrm{"}\,-2x-\frac{3}{2}
$$
for large negative $x$.
