#### Answer

$f(x) $ is decreasing on $(-\infty,-3/4) $ and increasing on $ ( -3/4,0) \cup (0 ,\infty) $.
$f(x)$ has a local minimum at $x= -3/4$.

#### Work Step by Step

Given $$y=x^{4}+x^{3}$$
Since
$$f'(x) =4x^{3}+3x^2 $$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
4x^{3}+3x^2 &=0\\
x^2(4x +3)&=0
\end{align*}
Then $x= 0$ and $x=-3/4$ are critical points. To find the intervals where $f$ is increasing and decreasing, choose $x=-1,\ x= -0.5$ and $x= 1$
\begin{align*}
f'( -1)&<0 \\
f'(-0.5)& >0\\
f'(1)&>0
\end{align*}
Hence, $f(x) $ is decreasing on $(-\infty,-3/4) $ and increasing on $ ( -3/4,0) \cup (0 ,\infty) $. Hence, by using the first derivative test, $f(x)$ has a local minimum at $x= -3/4$. $f(0)$ is a point of inflection.