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