\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{3}}}}}{{{9}}}}{\left({3}{\ln{{x}}}-{1}\right)}\)

Differentiate using the product rule

\(\displaystyle{f}'{\left({x}\right)}={\left({\frac{{{x}^{{{3}}}}}{{{9}}}}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{3}{\ln{{x}}}-{1}\right]}+{\left({3}{\ln{{x}}}-{1}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{x}^{{{3}}}}}{{{9}}}}\right]}\)

Therefore,

\(\displaystyle{f}'{\left({x}\right)}={\left({\frac{{{x}^{{{3}}}}}{{{9}}}}\right)}{\left({\frac{{{3}}}{{{x}}}}\right)}+{\left({3}{\ln{{x}}}-{1}\right)}{\left({\frac{{\lbrace}}{{3}}}{x}^{{{2}}}{\left\lbrace{9}\right\rbrace}\right)}\)

Simplify

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{3}}}}+{x}^{{{2}}}{\ln{{x}}}-{\frac{{{x}^{{{2}}}}}{{{3}}}}\)

\(\displaystyle{f}'{\left({x}\right)}={x}^{{{2}}}{\ln{{x}}}\)

Differentiate using the product rule

\(\displaystyle{f}'{\left({x}\right)}={\left({\frac{{{x}^{{{3}}}}}{{{9}}}}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{3}{\ln{{x}}}-{1}\right]}+{\left({3}{\ln{{x}}}-{1}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{x}^{{{3}}}}}{{{9}}}}\right]}\)

Therefore,

\(\displaystyle{f}'{\left({x}\right)}={\left({\frac{{{x}^{{{3}}}}}{{{9}}}}\right)}{\left({\frac{{{3}}}{{{x}}}}\right)}+{\left({3}{\ln{{x}}}-{1}\right)}{\left({\frac{{\lbrace}}{{3}}}{x}^{{{2}}}{\left\lbrace{9}\right\rbrace}\right)}\)

Simplify

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{3}}}}+{x}^{{{2}}}{\ln{{x}}}-{\frac{{{x}^{{{2}}}}}{{{3}}}}\)

\(\displaystyle{f}'{\left({x}\right)}={x}^{{{2}}}{\ln{{x}}}\)