Answer
(a) $x=0$ is a vertical asymptote.
(b) $f$ is decreasing on the intervals $(0, 1)\cup (2, \infty)$
$f$ is increasing on the interval $(1,2)$
(c) The local maximum is $f(2) = 0.87$
The local minimum is $f(1) = 0.83$
(d) The graph is concave down on this interval: $(\sqrt{2}, \infty)$
The graph is concave up on this interval: $(0, \sqrt{2})$
The point of inflection is $(1.41, 0.85)$
(e) We can see a sketch of the graph below.

Work Step by Step
(a) $f(x) = x -\frac{1}{6}x^2-\frac{2}{3}ln~x$
$f(x)$ is defined for values of $x$ such that $x \gt 0$
$x=0$ is a vertical asymptote.
$\lim\limits_{x \to \infty}f(x) = -\infty$
There is no horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = 1-\frac{1}{3}x-\frac{2}{3x} = 0$
$3x-x^2-2 = 0$
$x^2-3x+2 = 0$
$(x-2)(x-1) = 0$
$x = 1,2$
When $0 \lt x \lt 1$ or $x \gt 2$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(0, 1)\cup (2, \infty)$
When $1 \lt x \lt 2$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(1,2)$
(c) $f(1) = (1) -\frac{1}{6}(1)^2-\frac{2}{3}ln~(1) = \frac{5}{6} = 0.83$
$f(2) = (2) -\frac{1}{6}(2)^2-\frac{2}{3}ln~(2) = 0.87$
The local maximum is $f(2) = 0.87$
The local minimum is $f(1) = 0.83$
(d) We can find the points where $f''(x) = 0$:
$f''(x) = -\frac{1}{3}+\frac{2}{3x^2} = 0$
$\frac{1}{3} = \frac{2}{3x^2}$
$x^2 = 2$
$x = \sqrt{2} = 1.41$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(\sqrt{2}, \infty)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(0, \sqrt{2})$
$f(\sqrt{2}) = (\sqrt{2}) -\frac{1}{6}(\sqrt{2})^2-\frac{2}{3}ln~(\sqrt{2}) = 0.85$
The point of inflection is $(1.41, 0.85)$
(e) We can see a sketch of the graph below.
