Of the top of my head...
Definitely not this salt. There will be little free ammonia from the solvation of the $\ce{NH4^+}$ cation. The $\ce{Cl^-}$ is a spectator anion.
The $\ce{Na^+}$ is a spectator cation. The $\ce{CO3^{2-}}$ will be mostly $\ce{CO3^{2-}}$ with a tiny bit of $\ce{HCO3^-}$. So the ratio of $\frac{\ce{HCO3^-}}{\ce{CO3^{2-}}}$ (and hence the pH) will change rapidly with the addition of $\ce{H^+}$.
Some of ammonium will protonate the acetate anion. So you end up with a mixture of $\ce{NH4^+}$, $\ce{NH3}$, $\ce{CH3COO^-}$ and $\ce{CH3COOH}$. I think this solution would have the most buffer capacity for acid.
It seems you need to solve all of these for $\dfrac{d\text{pH}}{d\ce{H^+}}$ -- well the last two anyway...
SOLUTION
My calculus-foo is lost on this, so let's just brute force an answer. Let's assume we have 1.00 liters of 0.1 molar solutions of each of the salts. We'll calculate the pH for the salt, add $1.00\times10^{-3}$ moles of $\ce{H^+}$ and calculate the new pH. We can then calculate:
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}}$
$\ce{NH4^+ <=> H^+ + NH3 }$
$\text{K}_\text{a} = 5.75\times10^{-10} = \dfrac{\ce{[H^+][NH3]}}{\ce{[NH4^+]}}$
Salt alone
assume $\ce{[H^+]=[NH3]}$ and $\ce{[NH4^+]=0.1}$ molar then simplify
$\ce{[H^+]} = \sqrt{\text{K}_\text{a}\times\ce{[NH4^+]}} = 7.56\times10^{-6}$
pH = 5.121
we can now solve for
$\dfrac{\ce{[NH3]}}{\ce{[NH4^+]}} = \dfrac{\text{K}_\text{a}}{\ce{[H^+]}} = \dfrac{5.75\times10^{-10}}{7.56\times10^{-6}} = 7.61\times 10^{-5} $
$\ce{[NH3] = 7.61\times10^{-6}}$
Salt plus 0.001 moles acid
So there isn't any significant amount of $\ce{NH3}$ to protonate and the acid added can be added to the initial concentration of $\ce{H^+}$ from the salt.
final $\ce{[H^+]} = 7.56\times10^{-6} + 1.00\times10^{-3} = 1.008\times10^{-3}$
pH = 2.997
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{2.997 - 5.121 }{1.00\times10^{-3}} = -2124$
$\ce{CO3^{2-} + H2O <=> OH^- + HCO3^-}$
$\text{K}_\text{b} = 2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$
Salt alone
assume $x = \ce{[HCO3^-] = [OH^-]}$ and $\ce{[CO3^{2-}]} = 0.100 -x$ molar then simplify
$0 = x^2 + 2.14\times10^{-4}x - 2.14\times10^{-5}$
$x = 4.52\times10^{-3}$
$\ce{[OH^-]} = 4.52\times10^{-3}$
$\ce{[H^+]} = \dfrac{K_\rm{w}}{\ce{[OH^-]}} =2.21\times10^{-12}$
pH = 11.655
Salt plus 0.001 moles acid
Now if we add 0.001 moles of $\ce{H^+}$.
We'll essentially neutralize some $\ce{OH^-}$ and make some $\ce{HCO3^-}$, but we know that
$\ce{[HCO3^-] - [OH^-]} = 0.001$ or $\ce{[HCO3^-]} = 0.001 + \ce{[OH^-]}$
Let $x = \ce{[OH-]}$
$2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$
$2.14\times10^{-4} = \dfrac{(x)(0.001+x)}{0.1 - (0.001+x)} = \dfrac{x^2 + 0.001x}{0.099-x}$
$0 = x^2 + 1.214\times10^{-3}x - 2.1186\times10^{-4}$
$x = 0.004036$
$\ce{[OH^-]} = 0.004036$
$\ce{[H^+]} = \dfrac{1.00\times10^{-14}}{\ce{[OH^-]}} = 2.48\times10^{-12}$
pH = 11.606
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{11.655 - 11.606}{1.00\times10^{-3}} = 49$
Salt alone
Salt plus 0.001 moles acid