Setting $y=3^{x}$ in the given equation yields $$y^{2}+y=6 \Longrightarrow y^{2}+y-6=0 \Longrightarrow y=-3,2$$ Since $y>0$ we must have $$3^{x}=2 \Longrightarrow x=\log _{3}(2) \Longrightarrow 1 / x=\log _{2}(3)$$ This means that $$16^{1 / x}+4^{1 / x}=\left(2^{1 / x}\right)^{4}+\left(2^{1 / x}\right)^{2}=3^{4}+3^{2}=90$$