Answer
$f(\frac{2}{9})$ is a local minimum.
$f''(0)$ is not defined and the second derivative test can not be applied at the point $0$.
Work Step by Step
$f'(x)=9x^{1/2}-2x^{-1/2}$ is differentiable everywhere except at $0$. Therefore, $0$ is a critical point of $y$. The other critical point is the solution of $9x^{1/2}-2x^{-1/2}=0$.
$9x^{1/2}-2x^{-1/2}=0\implies9x^{1/2}=2x^{-1/2}$
$\implies \frac{2}{9}=x$.
The critical points are $\frac{2}{9}$ and $0$.
Now, $f''(0)=\frac{9}{2}(0)^{-1/2}+(0)^{-3/2}$ is not defined and the second derivative test can't be applied at the point $0$.
$f''(\frac{2}{9})=\frac{9}{2}(\frac{2}{9})^{-1/2}+(\frac{2}{9})^{-3/2}=positive$
According to the second derivative test, if $f'(\frac{2}{9})=0$ and $f''(\frac{2}{9})$ is positive, then $f(\frac{2}{9})$ is a local minimum.
