Answer
$\frac{e^x(2x^2-3x)}{(2x^2+x+1)^2}$
Work Step by Step
a) Applying the quotient rule, we get:
$$\frac{d}{dx}f(x)=\frac{(2x^2+x+1)(\frac{d}{dx}e^x)-e^x(\frac{d}{dx}2x^2+x+1)}{(2x^2+x+1)^2}$$
$$=\frac{(e^x)(2x^2+x+1)-e^x(4x+1)}{(2x^2+x+1)^2}$$
$$=\frac{e^x(2x^2-3x)}{(2x^2+x+1)^2}$$
b) See graph below (f is in red, f' is in blue).