Answer
$$ y'=x^{3^x}( 3^x\ln3\ln x+\frac{3^x}{x}).$$
Work Step by Step
Recall that $(e^x)'=e^x $, $(\ln x)'=\dfrac{1}{x}$. We have
$$ f(x)=x^{3^x}=e^{ 3^x\ln x}.$$ Now taking the derivative, we get $$ f'(x)= e^{ 3^x\ln x}( 3^x\ln x)'=e^{ 3x\ln x}( 3^x\ln3\ln x+\frac{3^x}{x})\\ =x^{3^x}( 3^x\ln3\ln x+\frac{3^x}{x}).$$
