Im trying to prove that $$e^z\le{1+z+\frac{\frac{z^2}{2}}{1-\frac{|z|}{3}}}$$ for $|z|<3$.
I tried looking at the expansion for the exponential function and have shown the bound for $|z|<1/2$ by considering:
$$e^z=1+z+\sum_{n=2}^{\infty}{\frac{z^n}{n!}}$$.
Prior to edit: The inequality $e^{z}\le 1+z+\dfrac{\dfrac{z^{\color{red}{1}}}{2}}{1-\frac{\left|z\right|}{3}}$ is untrue. Take $z=\pm2$. Then, $\require{enclose}e^{2}=7.389\enclose{updiagonalstrike}{\le}6=\left(1+2+\frac{\frac{2}{2}}{1-\frac{\left|2\right|}{3}}\right)$. In fact, the inequality is only true for $z\in[0,1.05\ldots].$
Post question edit: For nonpositive $x$, the inequality $e^{z}\le 1+z+\dfrac{\dfrac{z^{\color{red}{2}}}{2}}{1-\frac{\left|z\right|}{3}}$ can follow from the power series for $e^x$ with some manipulations.
$$e^x=1+x+\frac{x^2}{2}+\ldots \\ e^{x}=\frac{1}{e^{-x}}=\frac{1}{1-x+\frac{x^2}{2}+\ldots}\le \frac{1}{1-x+\frac{x^{2}}{2}} \\ e^{x}\le \frac{1}{1-x+\frac{x^{2}}{2}}\le^? 1+x+\frac{\frac{x^{2}}{2}}{1-\frac{\left|x\right|}{3}} $$
Expanding the last two rational functions in the final line gives the inequality $0\le 5x^{4}-2x^{3}$, which proves the result.
The case for positive $x$ is a bit trickier but you could consider the derivatives of $1+x+\frac{\frac{x^{2}}{2}}{1-\frac{x}{3}}-e^x$, the first of which is $1+\frac{18x-3x^{2}}{2\left(9-6x+x^{2}\right)}-e^{x}$ and the second is $-\left(\frac{27}{\left(x-3\right)^{3}}+e^{x}\right)$. So, the second derivative is positive if $e^{x}<\frac{27}{\left(3-x\right)^{3}}$, i.e., $\frac{x^{3}}{e^{x}}<\left(\frac{3}{e}\right)^{3}$ assuming $x\in(0,3)$. We can then see that $\frac{x^{3}}{e^{x}}$ is strictly positive for $x>0$, intersects $(0,0)$, and attains a single supremum when $\frac{e^{x}}{x^{3}}\left(1-\frac{3}{x}\right)=0$, i.e., $x=3$. Therefore, for $x\in(0,3)$, $0<\frac{x^{3}}{e^{x}}<\left(\frac{3}{e}\right)^{3}$ and the aforementioned second derivative is positive in the integral. So, the aforementioned first derivative is strictly increasing from $(0,0)$ and must be positive. Lastly, $1+x+\frac{\frac{x^{2}}{2}}{1-\frac{x}{3}}-e^x$ is increasing from $(0,0)$ and the inequality is true for $x\in(0,3)$. $\square$