f(x)= 3+(12/(x-3)) over 2-16(x/x^2-9)
OK, let me LaTeX it so I can read it to start with...
$\displaystyle f(x) = \frac{3+\frac{12}{x-3}}{2 - 16\frac{x}{x^2 - 9}}$
Let's get the top and bottom to have the same denominator to start with...
$\displaystyle =\frac{\frac{3(x-3)+12}{x-3}}{\frac{2(x^2-9)-16x}{x^2-9}}$
And I don't like having fractions in a numerator and denominator, so let's write this as a product instead...
$\displaystyle =\frac{3(x-3)+12}{x-3}\times \frac{x^2-9}{2(x^2-9)-16x}$
Expand...
$\displaystyle =\frac{3x+3}{x-3}\times \frac{x^2-9}{2x^2-18-16x}$
Factorise everything possible...
$\displaystyle \frac{3(x+1)}{x-3}\times \frac{(x+3)(x-3)}{2(x+1)(x-9)}$
Cancelling everything possible leaves...
$\displaystyle \frac{3(x+3)}{2(x-9)}$
There you go...