Answer
$$y'=(3x-2)^5-x^{-3}\Big(4-\frac{1}{2x^2}\Big)^{-2}$$
Work Step by Step
$$y=\frac{1}{18}(3x-2)^6+\Big(4-\frac{1}{2x^2}\Big)^{-1}$$
The derivative of function $y$ is: $$y'=\frac{1}{18}\Big[(3x-2)^6\Big]'+\Bigg[\Big(4-\frac{1}{2x^2}\Big)^{-1}\Bigg]'$$
$$y'=\frac{1}{18}\Big[6(3x-2)^5(3x-2)'\Big]+\Bigg[-\Big(4-\frac{1}{2x^2}\Big)^{-2}\Big(4-\frac{1}{2x^2}\Big)'\Bigg]$$
$$y'=\frac{1}{18}\Big[6(3x-2)^5\times3\Big]+\Bigg[-\Big(4-\frac{1}{2x^2}\Big)^{-2}\Big(0-\Big(\frac{-1(2x^2)'}{4x^4}\Big)\Big)\Bigg]$$
$$y'=\frac{1}{18}\Big[18(3x-2)^5\Big]+\Bigg[-\Big(4-\frac{1}{2x^2}\Big)^{-2}\Big(\frac{4x}{4x^4}\Big)\Bigg]$$
$$y'=(3x-2)^5-\Big(4-\frac{1}{2x^2}\Big)^{-2}\frac{1}{x^3}$$
$$y'=(3x-2)^5-x^{-3}\Big(4-\frac{1}{2x^2}\Big)^{-2}$$
