Derivative of a product involving an exponential and secant

By the product rule, $g'(x) = \frac{d}{dx}[e^{3x}]\cdot \sec(x) + e^{3x}\cdot \frac{d}{dx}[\sec(x)]$. The chain rule gives $\frac{d}{dx}[e^{3x}] = 3e^{3x}$ (outer function $e^u$, inner function $u=3x$), and $\frac{d}{dx}[\sec x] = \sec x \tan x$. So $g'(x) = 3e^{3x}\sec(x) + e^{3x}\sec(x)\tan(x)$.

Now evaluate at $x=0$: $\sec(0) = 1$ and $\tan(0) = 0$, and $e^{3\cdot 0} = e^0 = 1$. Substituting gives $g'(0) = 3(1)(1) + (1)(1)(0) = 3$.