Answer
$$e^{x^3}.$$
Work Step by Step
Since we have$$
1+x^{3}+\frac{x^{6}}{2 !}+\frac{x^{9}}{3 !}+\frac{x^{12}}{4 !}+\cdots\\
=1+x^{3}+\frac{(x^{3})^{2}}{2 !}+\frac{(x^{3})^{3}}{3 !}+\frac{(x^{3})^{4}}{4 !}+\cdots\\
$$
Then by using Table 2, we see that this is Maclaurin series of the function
$$e^{x^3}.$$
