differentiate

 1⁄3x^6(x^2-3)

$$\\y= 1/3x^6(x^2-3)\\\\ y=\frac{x^6}{3}\times (x^2-3)\\\\ y=\frac{x^8}{3}-x^6\\\\ \frac{dy}{dx}=\frac{8x^7}{3}-6x^5\\\\$$

1⁄3x^6(x^2-3) \\y= 1/3x^6(x^2-3)\\\\ y=\frac{x^6}{3}\times (x^2-3)\\\\ y=\frac{x^8}{3}-x^6\\\\ \frac{dy}{dx}=\frac{8x^7}{3}-6x^5\\\\ Like melody's answer