对 $z = e^{x+y} - \ln(x+y)$，先求一阶偏导数：
$$
\frac{\partial z}{\partial x} = e^{x+y} - \frac{1}{x+y},\quad
\frac{\partial z}{\partial y} = e^{x+y} - \frac{1}{x+y}.
$$
再求混合偏导数 $\frac{\partial^2 z}{\partial x \partial y}$，即对 $\frac{\partial z}{\partial x}$ 关于 $y$ 求偏导：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left(e^{x+y} - \frac{1}{x+y}\right)
= e^{x+y} + \frac{1}{(x+y)^2}.
$$
代入点 $(1,0)$，得 $x+y = 1$，故
$$
\left.\frac{\partial^2 z}{\partial x \partial y}\right|_{(1,0)} = e^{1} + \frac{1}{1^2} = e + 1.
$$
因此答案为 $e+1$。