Notice that
$$\frac{12x^3}{3x} = \frac{12}{3}\,\frac{x^3}{x} = 4x^2$$
so that you have $y=6x^4 - 4x^2$. Now use the fact that the derivative of a difference is the difference of the derivatives (if they both exist) and go from there.
If the original problem was
$$y = \frac{6x^4-12x^3}{3x},$$
then
$$y = \frac{6x^4}{3x} - \frac{12x^3}{3x} = 2x^3 - 4x^2;$$
given your comment, though, it seems you miscopied the problem and the $x^3$ should have been an $x^2$, i.e.,
$$y = \frac{6x^4 - 12x^2}{3x} = \frac{6x^4}{3x} - \frac{12x^2}{3x} = 2x^3 - 4x.$$
Then you can take derivatives.
The second problem, assuming it's
$$y = \frac{x^5+3x^3-2x^2}{x}$$
is solved the same way:
$$y = \frac{x^5 + 3x^3 - 2x^2}{x} = \frac{x(x^4+3x^2-2x)}{x} = x^4 +3x^2 - 2x.$$
$2x^3-4x$
$= 6x^2-4$ – 2011-09-29