Step 1: Argue from the definition of increasing directly, without calculus first, then confirm with the derivative:
Take any $x_1 < x_2$ in $R$. Since $e^t$ is a strictly increasing function of $t$, and $2x_1<2x_2$ (multiplying an inequality by the positive number 2 keeps its direction), it follows $e^{2x_1}<e^{2x_2}$, i.e. $f(x_1)<f(x_2)$.
Step 2: Confirm using the derivative test:
$f'(x)=2e^{2x}$. Since $e^{2x}>0$ always, $f'(x)>0$ for every real $x$.
Final Answer:
Both the direct argument and the derivative test confirm $f(x)=e^{2x}$ is strictly increasing on $R$.
\[ \boxed{f'(x)=2e^{2x}>0} \]