Can you do it without calculus?

$f(x)=9^{x}-3^{x}+1=3^{2x}-3^{x}+1=(3^x)^{2}-3^{x}+1=(3^{x}-\dfrac{1}{2})^{2}+\dfrac{3}{4}$ . Since $(3^{x}-\dfrac{1}{2})^{2} \ge 0$ , minimum value of $f(x)$ is $0+\dfrac{3}{4}=0.75$ and it occurs when $x=\log_{3}{0.5}$ .

Let's call $3^{x} = y$ then $9^{x} - 3^{x} + 1 = y^2 - y + 1$ . Let $f(y) = y^2 - y + 1$ , this is a parabola "up" with absolut minimum for $y = \frac{1}{2}$ ( $f '(y) = 2y - 1 =0$ if and only if $y =\frac{1}{2}$ ) $\Rightarrow$ the minimum is $f(\frac{1}{2})= \frac{1}{4} - \frac{1}{2} + 1 = \boxed{\frac{3}{4}}$

( $y = 3^{x} = \frac{1}{2} \Rightarrow x = \log_{3}{0.5} = \log_{3}{\frac{1}{2}}$ ) . Sorry, I didn't read the title.

$m=3^x \Rightarrow f(x)=9^x-3^x+1=m^2-m+1$ , minimum value of $f(x)$ is when $m=-\frac{1}{2} \therefore$ minimum value is $(\frac{1}{2})^2 - \frac{1}{2}+1=\boxed{\frac{3}{4}}$