Answer
$\left(\frac{-\ln 3}{2},\infty\right)$
Work Step by Step
The function $f$ is increasing on $I$ if $f'(x) \gt 0$ on $I$ so:
$$f'(x)=3e^{3x}-e^{x}=e^{x}(3e^{2x}-1)$$
since $e^{x} \gt 0$ for all $x \in \mathbb R$ it follows:
$$f'(x) \gt 0 \to (3e^{2x}-1)\gt 0$$
$$f'(x) \gt 0 \to 3e^{2x}\gt 1$$
$$f'(x) \gt 0 \to e^{2x}\gt \frac{1}{3}$$
$$f'(x) \gt 0 \to 2x\gt \ln\frac{1}{3}$$
$$f'(x) \gt 0 \to 2x\gt -\ln 3$$
$$f'(x) \gt 0 \to x\gt \frac{-\ln 3}{2}$$
so $f$ is increasing on $\left(\frac{-\ln 3}{2},\infty\right)$
