What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be minimum, it gave me an absurd relation like $-6=6$. 
How can we solve this? Why is the differentiation method not working here? Please help !!!
$f(x)= \frac{x^2+3x-6}{x^2+3x+6}$
$f'(x)= \frac{(x^2+3x+6)(2x+3)-(x^2+3x-6)(2x+3)}{(x^2+3x+6)^2}$
Then equated this to $0$ to find the value of $x$.
Thanks in advance !!!

Another didactic way, we prove that $-11/5 \le y=f(x)<1, x \in \Re$.
Let $y=\frac{x^2+3x-6}{x^2+3x+6}$ and check that $x^2+3x+6 >0$
Then we can write $(y-1)x^2+3(y-1)x+6(y+1)=0, x \in \Re$
$\implies B^2 \ge 4AC \implies 9(y-1)^2\ge 24(y+1)(y-1).$
Case 1: $y<1$ we get $9y-9 \le 24y+24 \implies  y \ge -11/5$
and hence $-11/5 \le y <1$.
Case 2: when $y>1$ gives $y\le -11/5$ a Null.
