Answer
$y = -3x+1~~$ is the slant asymptote.
Work Step by Step
$y = \frac{-6x^4+2x^3+3}{2x^3-x}$
We can express $-6x^4+2x^3+3$ in terms of $2x^3-x$:
$-6x^4+2x^3+3 = (2x^3-x)(-3x+1)+(-3x^2+x+3)$
Then:
$\frac{-6x^4+2x^3+3}{2x^3-x} = (-3x+1)+\frac{-3x^2+x+3}{2x^3-x}$
We can see that $~~y = -3x+1~~$ is a good candidate for the slant asymptote.
We can evaluate the limit as $x \to -\infty$:
$\lim\limits_{x \to -\infty} [\frac{-6x^4+2x^3+3}{2x^3-x}-(-3x+1)]$
$= \lim\limits_{x \to -\infty} [\frac{-6x^4+2x^3+3}{2x^3-x}-\frac{(-3x+1)(2x^3-x)}{2x^3-x}]$
$= \lim\limits_{x \to -\infty} (\frac{-6x^4+2x^3+3}{2x^3-x}-\frac{-6x^4+2x^3+3x^2-x}{2x^3-x})$
$= \lim\limits_{x \to -\infty} \frac{-3x^2+x+3}{2x^3-x}$
$= \lim\limits_{x \to -\infty} \frac{-3x^2/x^3+x/x^3+3/x^3}{2x^3/x^3-x/x^3}$
$= \lim\limits_{x \to -\infty} \frac{-3/x+1/x^2+3/x^3}{2-1/x^2}$
$= \frac{0+0+0}{2-0}$
$= 0$
We can evaluate the limit as $x \to \infty$:
$\lim\limits_{x \to \infty} [\frac{-6x^4+2x^3+3}{2x^3-x}-(-3x+1)]$
$= \lim\limits_{x \to \infty} [\frac{-6x^4+2x^3+3}{2x^3-x}-\frac{(-3x+1)(2x^3-x)}{2x^3-x}]$
$= \lim\limits_{x \to \infty} (\frac{-6x^4+2x^3+3}{2x^3-x}-\frac{-6x^4+2x^3+3x^2-x}{2x^3-x})$
$= \lim\limits_{x \to \infty} \frac{-3x^2+x+3}{2x^3-x}$
$= \lim\limits_{x \to \infty} \frac{-3x^2/x^3+x/x^3+3/x^3}{2x^3/x^3-x/x^3}$
$= \lim\limits_{x \to \infty} \frac{-3/x+1/x^2+3/x^3}{2-1/x^2}$
$= \frac{0+0+0}{2-0}$
$= 0$
Therefore, $~~y = -3x+1~~$ is the slant asymptote.