Answer
$$ e^{3x} = 1+\dfrac{1}{1!}(3x)+\dfrac{1}{2!}(3x)^2+\dfrac{1}{3!}(3x)^3+.....+\dfrac{3^n}{n!}x^n $$
Work Step by Step
Since
\begin{align*}
f(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(0)=1\\
f'(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ f'(0)=1\\
f''(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ f''(0)=1\\
f'''(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ f'''(0)=1\\
f^{(4)}(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ f^{(4)}(0)= 1
\end{align*}
Then
\begin{align*}
e^x &=f(0)+\dfrac{f'(0)}{1!}x+\dfrac{f''(0)}{2!}x^2+\dfrac{f'''(0)}{3!}x^3+.....+\dfrac{f^{(n)}(0)}{n!}x^n\\
&= 1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^2+\dfrac{1}{3!}x^3+.....+\dfrac{1}{n!}x^n
\end{align*}
To find the Maclaurin series of $f(x)=e^{3x}$, replace $x$ by $3x$:
$$ e^{3x} = 1+\dfrac{1}{1!}(3x)+\dfrac{1}{2!}(3x)^2+\dfrac{1}{3!}(3x)^3+.....+\dfrac{3^n}{n!}x^n $$
