f(x)=1/(3*(x+2)^{2}(x-3)^{2}). Since we have squares of (x-3) and (x+2), the function is non-negative everywhere. When x goes to ±∞, function tends to zero, so y=0 is horizontal asymptotic line. Critical points are x=-2 and x=3, those are the points where denominator equals zero. So x=-2 and x=3 are vertical asymptotic lines.

f'(x)=-2/3*(2x-1)/[(x+2)(x-3)]^{3}; f'(x)=0 at x=½.

f''(x)=10/3*(2x^{2}-2x+3)/[(x+2)(x-3)]^{4};

f''(x) is positive everywhere. The function is concave everywhere.

Function increases on the intervals: (-∞, -2); (½, 3)

Function decreases on the intervals: (-2,½); (3,∞)