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: $y = \frac{x^5 + 3x^3 - 2x^2}{x} = \frac{x(x^4+3x^2-2x)}{x} = x^4 +3x^2 - 2x.$