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