jmc

Let $y=3^x.$ Then $$9^x-3^x+1=y^2-y+1={\left(y-\frac{1}{2}\right)}^2+\frac{3}{4}.$$Thus, the minimum value is $\boxed{\frac{3}{4}},$ which occurs when $y=\frac{1}{2},$ or $x={\log }_3\frac{1}{2}.$