Answer
$$y'=\frac{3}{x\ln4}$$
Work Step by Step
$$y=\log_4x+\log_4x^2$$
We have $\log_ax+\log_ay=\log_a(xy)$. Therefore, $$y=\log_4(x\times x^2)=\log_4x^3$$
Recall Theorem 7: $$\frac{d}{dx}\log_au=\frac{1}{u\ln a}\frac{du}{dx}$$
Apply here to find $y'$: $$y'=(\log_4x^3)'=\frac{1}{x^3\ln4}(x^3)'=\frac{3x^2}{x^3\ln4}$$
$$y'=\frac{3}{x\ln4}$$
