How to find the range of the given function? 
Find the range of $$f(x)=\dfrac{x^2+14x+9}{x^2+2x+3}$$ where $x\in \mathbb R$ 

I thought of finding derivative but this will get too complicated so i am completely blank.
Thanks in advance!

Just some general notes on finding the range for a rational function:
First of all a rational function is continuous where it's defined so if $y_1$ and $y_2$ is in the range then so is any value inbetween $y_1$ and $y_2$ (the intermediate value theorem). This is a key fact needed to determine the range.
First we check that the denominator is not zero. If the denominator has a zero at $x=x_*$ then $\lim_{x\to x_*}|f(x)| = \infty$ so the range will extend to $\infty$ or $-\infty$ (or both). You can determine this by computing the one-sided limits $\lim_{x\to x_*^{\pm}} f(x)$. For your function we have $x^2+2x+3 = (x+1)^2+2 \geq 2$ so this is not the case here.
Next we determine the asymptotic values by computing $\lim_{x\to\pm\infty} f(x)$. Again if one or both of these limits are $\pm\infty$ then the range will exend out to these values. For your case these limits are $1$ and togeather with the previous information we can conclude that the function is bounded.
In after any of the steps above if the range has beens shown to be $\mathbb{R}$ then we can stop. But if this is not the case then there is one more step needed. We need to determine the local max/min values of the function. This is done by solving $f'(x) = 0$ and computing $f(x)$ using the values found. For your function we find the extremal points $x=-2$ and $x=1$ and $f(-2) = -5$ and $f(1) = 4$. The point $x=-2$ is a global minimum and $x=1$ is a global maximum.
Combinding all the information found by going through the steps above you should be able to determine the range quite easily. For your function it's not hard to see that the range is $[-5,4]$. For this last step it's often quite useful to draw a sketch based on the information previously found. For example I drew up this rough sketch:

