We need to determine the sign of$f'(x)=e^{x}\left(x^{2}+x1\right)$.
From part (e) we have $f'(x)=e^{x}\left(x^{2}+x1\right)$, where $e^{x}$ is always positive so it doesn't have any role in the sign. We only need to determine the sign of $x^{2}x+1$.
From part (f) $x^{2}x+1$ is $0\ \quad$ at $x={\dfrac {1+{\sqrt {5}}}{2}}\approx 0.35,\ {\text{and}}\ \ x={\dfrac {1{\sqrt {5}}}{2}}\approx 0.13$,
so we make a sign chart for $f'(x)$ to find where it is positive and negative and therefore where the function $f(x)$ is increasing and decreasing.

$\displaystyle (\infty ,{\dfrac {1{\sqrt {5}}}{2}})$ 
$\displaystyle ({\dfrac {1{\sqrt {5}}}{2}},{\dfrac {1+{\sqrt {5}}}{2}})$ 
$\displaystyle ({\dfrac {1+{\sqrt {5}}}{2}},\infty )$

$x^{2}x+1$ 
$\displaystyle $ 
$\displaystyle +$ 
$\displaystyle $

$f'(x)=e^{x}\left(x^{2}x+1\right)$ 
$\displaystyle $ 
$\displaystyle +$ 
$\displaystyle $

$f(x)$ 
decreasing 
increasing 
decreasing

Intervals that $f(x)$ is decreasing: ${\color {blue}(\infty ,{\dfrac {1{\sqrt {5}}}{2}}),\ ({\dfrac {1+{\sqrt {5}}}{2}},\infty )}$.
Intervals that $f(x)$ is increasing: ${\color {blue}({\dfrac {1{\sqrt {5}}}{2}},{\dfrac {1+{\sqrt {5}}}{2}})}$.