Answer
$f'(x)=\frac{-2x^2+2x-3}{(x^2-3x)^2}$
Work Step by Step
$f(x)=\frac{\frac{2x}{x}-\frac{1}{x}}{x-3}=\frac{2x-1}{x(x-3)}=\frac{2x-1}{x^2-3x}$
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=2x-1; u'(x)=2$
$v(x)=x^2-3x; v'(x)=2x-3$
$f'(x)=\frac{(2)(x^2-3x)-(2x-3)(2x-1)}{(x^2-3x)^2}$
$=\frac{2x^2-6x-(4x^2-8x+3)}{(x^2-3x)^2}$
$=\frac{-2x^2+2x-3}{(x^2-3x)^2}$
