Simplify f(z)

\(\displaystyle{f{{\left({z}\right)}}}={\frac{{{z}^{{{2}}}+{1}}}{{{3}{z}}}}={\frac{{{1}}}{{{3}}}}{\left({z}+{z}^{{-{1}}}\right)}\)

Use the power rule tog et

\(\displaystyle{f}'{\left({z}\right)}={\frac{{{1}}}{{{3}}}}{\left({1}-{z}^{{-{2}}}\right)}={\frac{{{z}^{{{2}}}-{1}}}{{{3}{z}^{{{2}}}}}}\)

\(\displaystyle{f{{\left({z}\right)}}}={\frac{{{z}^{{{2}}}+{1}}}{{{3}{z}}}}={\frac{{{1}}}{{{3}}}}{\left({z}+{z}^{{-{1}}}\right)}\)

Use the power rule tog et

\(\displaystyle{f}'{\left({z}\right)}={\frac{{{1}}}{{{3}}}}{\left({1}-{z}^{{-{2}}}\right)}={\frac{{{z}^{{{2}}}-{1}}}{{{3}{z}^{{{2}}}}}}\)