Given:

\(\displaystyle{9}^{{x}}-{3}^{{{x}+{1}}}+{1}={0}\)

\(\displaystyle{3}^{{{2}{x}}}-{3}^{{{x}}}\times{3}+{1}={0}\)

Let \(\displaystyle{3}^{{x}}\) be u:

\(\displaystyle{u}^{{2}}-{u}\times{3}+{1}={0}\)

\(\displaystyle{u}=\frac{{{3}+\sqrt{{5}}}}{{2}},{u}=\frac{{{3}-\sqrt{{5}}}}{{2}}\)

Substitute back \(\displaystyle{3}^{{x}}\):

\(\displaystyle{3}^{{x}}=\frac{{{3}+\sqrt{{5}}}}{{2}}\)

\(\displaystyle{\ln{{\left({3}^{{x}}\right)}}}={\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}\)

\(\displaystyle{x}{\ln{{\left({3}\right)}}}={\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}\)

\(\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}\)

Solve \(\displaystyle{3}^{{x}}=\frac{{{3}-\sqrt{{5}}}}{{2}}\) similarly:

\(\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}-\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}\)

\(\displaystyle{9}^{{x}}-{3}^{{{x}+{1}}}+{1}={0}\)

\(\displaystyle{3}^{{{2}{x}}}-{3}^{{{x}}}\times{3}+{1}={0}\)

Let \(\displaystyle{3}^{{x}}\) be u:

\(\displaystyle{u}^{{2}}-{u}\times{3}+{1}={0}\)

\(\displaystyle{u}=\frac{{{3}+\sqrt{{5}}}}{{2}},{u}=\frac{{{3}-\sqrt{{5}}}}{{2}}\)

Substitute back \(\displaystyle{3}^{{x}}\):

\(\displaystyle{3}^{{x}}=\frac{{{3}+\sqrt{{5}}}}{{2}}\)

\(\displaystyle{\ln{{\left({3}^{{x}}\right)}}}={\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}\)

\(\displaystyle{x}{\ln{{\left({3}\right)}}}={\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}\)

\(\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}+\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}\)

Solve \(\displaystyle{3}^{{x}}=\frac{{{3}-\sqrt{{5}}}}{{2}}\) similarly:

\(\displaystyle{x}=\frac{{{\ln{{\left(\frac{{{3}-\sqrt{{5}}}}{{2}}\right)}}}}}{{{\ln{{\left({3}\right)}}}}}\)