Remarks: The optimal $c$ is given by $c_0 = \max_{x\ne 0} \frac{\ln(\mathrm{e}^x - x)}{x^2}$. I think that it does not admit a closed form. Here we give a proof of $c = 9/16$.
Letting $x = \frac43 y$, the desired inequality is written as
$$\mathrm{e}^{y^2} + \frac43y \ge \mathrm{e}^{4y/3}. \tag{1}$$
Case 1: $y \ge 4/3$
Clearly (1) is true.
$\phantom{2}$
Case 2: $y < 4/3$
Using $\mathrm{e}^u \ge 1 + u + \frac12u^2 + \frac16u^3$ for all $u\ge 0$, we have
\begin{align*}
\mathrm{e}^{y^2} &\ge 1 + y^2 + \frac12y^4 + \frac16 y^6\\[6pt]
&\ge 1 + y^2 + \frac12y^4 + \frac16 \cdot \left(\frac12 y^4 - \frac{1}{16}y^2\right) \tag{2}\\[6pt]
&= \frac{7}{12}y^4 + \frac{95}{96}y^2 + 1\\[6pt]
&\ge \frac{33}{35}\cdot \frac{7}{12}y^4 + \frac{95}{96}y^2 + 1\\[6pt]
&= \frac{11}{20}y^4 + \frac{95}{96}y^2 + 1
\end{align*}
where in (2) we have used $y^6 + \frac{1}{16}y^2 \ge \frac12 y^4$ (AM-GM).
Thus, it suffices to prove that
$$\left(\frac{11}{20}y^4 + \frac{95}{96}y^2 + 1\right) + \frac43y
\ge \mathrm{e}^{4y/3}.$$
Since $\frac{11}{20}y^4 + \frac{95}{96}y^2 + \frac43y + 1
> 0$, it suffices to prove that
$$\ln\left(\frac{11}{20}y^4 + \frac{95}{96}y^2 + \frac43y + 1\right) \ge \frac{4y}{3}.$$
Denote $\mathrm{LHS} - \mathrm{RHS}$ by $f(y)$. We have
$$f'(y) = \frac{-1056y(y - a)(y- 1/2)(y - b)}{792y^4 + 1425y^2 + 1920y + 1440}$$
where
$$a = \frac54 - \frac{1}{132}\sqrt{17655} \approx 0.2434, \quad b = \frac54 + \frac{1}{132}\sqrt{17655} \approx 2.2566.$$
Thus, we have
$f'(y) < 0$ on $(-\infty, 0)$,
and $f'(y) > 0$ on $(0, a)$,
and $f'(y) < 0$ on $(a, 1/2)$,
and $f'(y) > 0$ on $(1/2, 4/3)$.
Also, we have $f(0) = 0$
and $f(1/2) = \ln\frac{1247}{640} - \frac23 > 0$. Thus, $f(y)\ge 0$
for all $y < 4/3$.
We are done.
