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Your input: find $$$\frac{\partial^{2}}{\partial x \partial y}\left(e^{x y}\right)$$$

Write the function $$$e^{x y}$$$ as a composition of the two functions $$$u=g=x y$$$ and $$$f\left(u\right)=e^{u}$$$.

Apply the chain rule $$$\frac{\partial}{\partial x} \left(f\left(g\right) \right)=\frac{\partial}{\partial u} \left(f\left(u\right) \right) \cdot \frac{\partial}{\partial x} \left(g \right)$$$:

$${\color{red}{\frac{\partial}{\partial x}\left(e^{x y}\right)}}={\color{red}{\frac{\partial}{\partial u}\left(e^{u}\right) \frac{\partial}{\partial x}\left(x y\right)}}$$

The derivative of an exponential is $$$\frac{\partial}{\partial u} \left(e^{u} \right)=e^{u}$$$:

$${\color{red}{\frac{\partial}{\partial u}\left(e^{u}\right)}} \frac{\partial}{\partial x}\left(x y\right)={\color{red}{e^{u}}} \frac{\partial}{\partial x}\left(x y\right)$$

Return to the old variable:

$$e^{{\color{red}{u}}} \frac{\partial}{\partial x}\left(x y\right)=e^{{\color{red}{x y}}} \frac{\partial}{\partial x}\left(x y\right)$$

Apply the constant multiple rule $$$\frac{\partial}{\partial x} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial x} \left(f \right)$$$ with $$$c=y$$$ and $$$f=x$$$:

$$e^{x y} {\color{red}{\frac{\partial}{\partial x}\left(x y\right)}}=e^{x y} {\color{red}{y \frac{\partial}{\partial x}\left(x\right)}}$$

Apply the power rule $$$\frac{\partial}{\partial x} \left(x^{n} \right)=n\cdot x^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial x} \left(x \right)=1$$$:

$$y e^{x y} {\color{red}{\frac{\partial}{\partial x}\left(x\right)}}=y e^{x y} {\color{red}{1}}$$

Thus, $$$\frac{\partial}{\partial x}\left(e^{x y}\right)=y e^{x y}$$$

Apply the product rule $$$\frac{\partial}{\partial y} \left(f \cdot g \right)=\frac{\partial}{\partial y} \left(f \right) \cdot g + f \cdot \frac{\partial}{\partial y} \left(g \right)$$$ with $$$f=y$$$ and $$$g=e^{x y}$$$:

$${\color{red}{\frac{\partial}{\partial y}\left(y e^{x y}\right)}}={\color{red}{\left(y \frac{\partial}{\partial y}\left(e^{x y}\right) + \frac{\partial}{\partial y}\left(y\right) e^{x y}\right)}}$$

Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial y} \left(y \right)=1$$$:

$$y \frac{\partial}{\partial y}\left(e^{x y}\right) + e^{x y} {\color{red}{\frac{\partial}{\partial y}\left(y\right)}}=y \frac{\partial}{\partial y}\left(e^{x y}\right) + e^{x y} {\color{red}{1}}$$

Write the function $$$e^{x y}$$$ as a composition of the two functions $$$u=g=x y$$$ and $$$f\left(u\right)=e^{u}$$$.

Apply the chain rule $$$\frac{\partial}{\partial y} \left(f\left(g\right) \right)=\frac{\partial}{\partial u} \left(f\left(u\right) \right) \cdot \frac{\partial}{\partial y} \left(g \right)$$$:

$$y {\color{red}{\frac{\partial}{\partial y}\left(e^{x y}\right)}} + e^{x y}=y {\color{red}{\frac{\partial}{\partial u}\left(e^{u}\right) \frac{\partial}{\partial y}\left(x y\right)}} + e^{x y}$$

The derivative of an exponential is $$$\frac{\partial}{\partial u} \left(e^{u} \right)=e^{u}$$$:

$$y {\color{red}{\frac{\partial}{\partial u}\left(e^{u}\right)}} \frac{\partial}{\partial y}\left(x y\right) + e^{x y}=y {\color{red}{e^{u}}} \frac{\partial}{\partial y}\left(x y\right) + e^{x y}$$

Return to the old variable:

$$y e^{{\color{red}{u}}} \frac{\partial}{\partial y}\left(x y\right) + e^{x y}=y e^{{\color{red}{x y}}} \frac{\partial}{\partial y}\left(x y\right) + e^{x y}$$

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=x$$$ and $$$f=y$$$:

$$y e^{x y} {\color{red}{\frac{\partial}{\partial y}\left(x y\right)}} + e^{x y}=y e^{x y} {\color{red}{x \frac{\partial}{\partial y}\left(y\right)}} + e^{x y}$$

Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial y} \left(y \right)=1$$$:

$$x y e^{x y} {\color{red}{\frac{\partial}{\partial y}\left(y\right)}} + e^{x y}=x y e^{x y} {\color{red}{1}} + e^{x y}=\left(x y + 1\right) e^{x y}$$

Thus, $$$\frac{\partial}{\partial y}\left(y e^{x y}\right)=\left(x y + 1\right) e^{x y}$$$

Therefore, $$$\frac{\partial^{2}}{\partial x \partial y}\left(e^{x y}\right)=\left(x y + 1\right) e^{x y}$$$

Answer: $$$\frac{\partial^{2}}{\partial x \partial y}\left(e^{x y}\right)=\left(x y + 1\right) e^{x y}$$$