Answer
\[\begin{align}
& \text{Decreasing on }\left( 0,1 \right),\left( 2,\infty \right) \\
& \text{Increasing on }\left( -\infty ,0 \right),\left( 1,2 \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=-\frac{{{x}^{4}}}{4}+{{x}^{3}}-{{x}^{2}} \\
& \text{Diferentiate } \\
& f'\left( x \right)=-\frac{4{{x}^{3}}}{4}+3{{x}^{2}}-2x \\
& f'\left( x \right)=-{{x}^{3}}+3{{x}^{2}}-2x \\
& \text{Calculate the critical points, set }f'\left( x \right)=0 \\
& -{{x}^{3}}+3{{x}^{2}}-2x=0 \\
& \text{Factoring} \\
& -x\left( {{x}^{2}}-3x+2 \right)=0 \\
& -x\left( x-2 \right)\left( x-1 \right)=0 \\
& \text{We obtain} \\
& x=0,\text{ }x=1,\text{ }x=2 \\
& \text{From the critical values we can make the following intervals}\, \\
& \left( -\infty ,0 \right),\left( 0,1 \right),\left( 1,2 \right),\left( 2,\infty \right) \\
& \text{Now, we will evaluate between the critical values and resume } \\
& \text{in a table} \\
& \text{ }\begin{matrix}
\text{Interval} & \text{Test value}\left( x \right) & \text{Sign of }{f}'\left( x \right) & \text{Behavior of }f\left( x \right) \\
\left( -\infty ,0 \right) & -1 & + & \text{Increasing} \\
\left( 0,1 \right) & 0.5 & - & \text{Decreasing} \\
\left( 1,2 \right) & 1.5 & + & \text{Increasing} \\
\left( 2,\infty \right) & 3 & - & \text{Decreasing} \\
{} & {} & {} & {} \\
{} & {} & {} & {} \\
\end{matrix} \\
& \text{From the table we can conlude that the function is:} \\
& \text{Decreasing on }\left( 0,1 \right),\left( 2,\infty \right) \\
& \text{Increasing on }\left( -\infty ,0 \right),\left( 1,2 \right) \\
& \text{Graph} \\
\end{align}\]