$\displaystyle f(x)=\frac{10x+3}{5x^2+3x-6}$

$\displaystyle f(x)=\frac{[(10)(5x^2+3x-6)]-[10x+3)(10x+3]}{(5x^2+3x-6)}$

$\displaystyle f(x)=\frac{(50x^2+30x-60)-(10x+3)^2}{(5x^2+3x-6)^2}$

$\displaystyle f(x)=\frac{(50x^2+30x-60)-(10x^2+9)}{(5x^2+3x-6)^2}$

$\displaystyle f'(x)= \frac{40x^2+30x-69}{(5x^2+3x-6)^2}$

Did i do that right or is the end part

$\displaystyle \frac{40x^2+30x-51}{(5x^2+3x-6)^2}$

????