By applying the Product Rule twice, one can prove that if $f$, $g$, and $h$ are differentiable, then $(fgh)'=f'gh+fg'h+fgh'$.
Now, in the above result, letting $f = g = h$ yields $\frac{d}{dx} [f(x)]^3 = 3[f(x)]^2 f'(x)$.
Use this last formula to differentiate $y=e^{3x}$.

$y'=$