Answer
$$y'=\frac{1}{10}\Big(\frac{3}{3x+4}-\frac{1}{x-2}\Big)\sqrt[10]{\frac{3x+4}{2x-4}}$$
Work Step by Step
$$y=\sqrt[10]{\frac{3x+4}{2x-4}}=\Big(\frac{3x+4}{2x-4}\Big)^{1/10}$$
Take the natural logarithm of both sides: $$\ln y=\ln\Big(\frac{3x+4}{2x-4}\Big)^{1/10}=\frac{1}{10}\ln\frac{3x+4}{2x-4}$$ $$\ln y =\frac{1}{10}(\ln(3x+4)-\ln(2x-4))$$
Now take the derivative of both sides, and remember that $(\ln x)'=1/x$:
$$\frac{1}{y}\times y'=\frac{1}{10}\Big(\frac{1}{3x+4}\times(3x+4)'-\frac{1}{2x-4}\times(2x-4)'\Big)$$ $$\frac{y'}{y}=\frac{1}{10}\Big(\frac{3}{3x+4}-\frac{2}{2x-4}\Big)$$ $$\frac{y'}{y}=\frac{1}{10}\Big(\frac{3}{3x+4}-\frac{1}{x-2}\Big)$$
Find $y'$: $$y'=\frac{1}{10}\Big(\frac{3}{3x+4}-\frac{1}{x-2}\Big)y$$ $$y'=\frac{1}{10}\Big(\frac{3}{3x+4}-\frac{1}{x-2}\Big)\sqrt[10]{\frac{3x+4}{2x-4}}$$
