\displaystyle f(x)=\frac{e^{x}}{1+x+\frac{1}{2}x^{2}+\frac{x^{3}}{6}}\\
=f(x)+xf(x)+\frac{1}{2}x^{2}f(x)+\frac{x^{3}}{6}f(x)=e^{x}\\
\sum_{n\geq 0}a_{n}x^{n}
+x\sum_{n\geq 1}a_{n-1}x^{n-1}
+\frac{1}{2}x^{2}\sum_{n\geq 2}a_{n-2}x^{n-2}
+\frac{x^{3}}{6}\sum_{n\geq 3}a_{n-3}x^{n-3}=e^{x}\\
\sum_{n\geq 0}a_{n}x^{n}+\sum_{n\geq 1}a_{n-1}x^{n}+\sum_{n\geq 2}\frac{1}{2}a_{n-2}x^{n}+\sum_{n\geq 3}\frac{1}{6}a_{n-3}x^{n}=e^{x}\\
a_{0}+a_{1}x+a_{2}x^{2}+\sum_{n\geq 3}a_{n}x^{n}+
a_{0}x+a_{1}x^{2}+\sum_{n\geq 3}a_{n-1}x^{n}+
\frac{1}{2}a_{0}x^{2}+\sum_{n\geq 3}\frac{1}{2}a_{n-2}x^{n}+\sum_{n\geq 3}\frac{a_{n-3}}{6}x^{n}=e^{x}\\
a_{0}+(a_{0}+a_{1})x+(\frac{1}{2}a_{0}+a_{1}+a_{2})x^{2}+
\sum_{n\geq 3}(a_{n}+a_{n-1}+\frac{1}{2}a_{n-2}+\frac{a_{n-3}}{6})x^{n}=e^{x}