Aloha :)
$$f(x)=\frac{3}{\sqrt[3]x}=\frac{3}{x^{\frac{1}{3}}}=3x^{-\frac{1}{3}}$$$$f'(x)=3\cdot\left(-\frac{1}{3}\right)x^{-\frac{4}{3}}=-x^{-\frac{4}{3}}=-\frac{1}{x^{\frac{4}{3}}}=-\frac{1}{x\sqrt[3]x}$$$$f''(x)=-\left(-\frac{4}{3}\right)x^{-\frac{7}{3}}=\frac{4}{3}x^{-\frac{7}{3}}=\frac{4}{3x^2\sqrt[3]x}$$
$$f(x)=3-\frac{\ln(x)}{3}+\frac{1}{\sqrt x}=3-\frac{\ln(x)}{3}+x^{-\frac{1}{2}}$$$$f'(x)=-\frac{1}{3}\,\frac{1}{x}-\frac{1}{2}x^{-\frac{3}{2}}=-\frac{1}{3}x^{-1}-\frac{1}{2}x^{-\frac{3}{2}}=-\frac{1}{3x}-\frac{1}{2x\sqrt x}$$$$f''(x)=-\frac{1}{3}(-1)x^{-2}+\frac{1}{2}\cdot\frac{3}{2}x^{-\frac{5}{2}}=\frac{1}{3x^2}+\frac{3}{4}x^{-\frac{5}{2}}=\frac{1}{3x^2}+\frac{3}{4x^2\sqrt x}$$