You need to identify if the function is odd or even such that:

`f(-x) = (-3x)/((-x)^2 - 4) =gt f(-x) = -(3x)/(x^2-4) = -f(x)`

Since `f(-x) = -f(x),` then the function is odd and the graph of an odd function has symmetry with respect to origin.

You need to identify the maximum and minimum of the function using the equation f'(x) = 0 such that:

`f'(x) = ((3x)'*(x^2-4) - (3x)*(x^2-4)')/((x^2-4)^2)`

`f'(x) = (3*(x^2-4) - 6x^2)/((x^2-4)^2)`

`f'(x) = (-3x^2 - 12)/((x^2-4)^2)`

You need to solve the equation `f'(x) = 0` => `3x^2 + 12 = 0`

`x^2 + 4 = 0` contradiction (there is no such a real value for x)

Hence, the function has no critical points and it decreases all over.

You need to identify inflection points, hence you need to solve the equation f''(x) = 0 such that:

`f''(x) = ((-3x^2 - 12)'*((x^2-4)^2) - (-3x^2 - 12)*((x^2-4)^2)')/((x^2-4)^4)`

`f''(x) = ((-6x)*((x^2-4)^2) - 4x(-3x^2 - 12)*(x^2-4))/((x^2-4)^4)`

`f''(x) = -2x(x^2-4)(3x^2-12-6x^2-24)/((x^2-4)^4)`

`f''(x) = -2x(-3x^2-36)/((x^2-4)^3)`

You need to solve the equation `f''(x) = 0` such that:

`-2x(-3x^2-36) = 0`

`-2x = 0 =gt x = 0`

`3x^2 + 36 = 0` contradiction

Hence, the function has an inflection point at `x = 0` .

You may find if the function has vertical asymptotes solving the equation `x^2 - 4 = 0` .

Notice that the function has vertical asymptotes at `x = +-2.`

You need to sketch the graph of function such that: