Answer
\begin{aligned} \\ \frac{\partial z}{\partial x} &=2 e^{y}+3 y e^{-x} \\
\end{aligned}\begin{aligned}
\frac{\partial^{2} z}{\partial x^{2}} &= -3 y e^{-x} \\
\end{aligned}\begin{aligned} \frac{\partial^{2} z}{\partial y \partial x} &=2 e^{y}+3 e^{-x} \\
\end{aligned} \begin{aligned} \frac{\partial z}{\partial y } &=2 x e^{y}-3 e^{-x} \\
\end{aligned}\begin{aligned} \frac{\partial^{2} z}{\partial y^{2}} &=2 x e^{y}
\end{aligned}\begin{aligned} \frac{\partial^{2} z}{\partial x \partial y} &=2 e^{y}+3 e^{-x}\end{aligned}
and we note
$$ \frac{\partial^{2} z}{\partial x \partial y}= \frac{\partial^{2} z}{\partial y \partial x}$$
Work Step by Step
Given $$z= 2 x e^{y}-3 y e^{-x}$$
The partial derivative with respect to $x$ is:
\begin{aligned} \\ \frac{\partial z}{\partial x} &=2 e^{y}+3 y e^{-x} \\
\end{aligned}
The second partial derivative with respect to $x$ is:
\begin{aligned}
\frac{\partial^{2} z}{\partial x^{2}} &= -3 y e^{-x} \\
\end{aligned}
\begin{aligned} \frac{\partial^{2} z}{\partial y \partial x} &=2 e^{y}+3 e^{-x} \\
\end{aligned}
The partial derivative with respect to $y$ is:
\begin{aligned} \frac{\partial z}{\partial y } &=2 x e^{y}-3 e^{-x} \\
\end{aligned}
The second partial derivative with respect to $y$ is:
\begin{aligned} \frac{\partial^{2} z}{\partial y^{2}} &=2 x e^{y}
\end{aligned}
\begin{aligned} \frac{\partial^{2} z}{\partial x \partial y} &=2 e^{y}+3 e^{-x}\end{aligned}
We see that the second mixed partial derivatives are equal.
$$ \frac{\partial^{2} z}{\partial x \partial y}= \frac{\partial^{2} z}{\partial y \partial x}$$
