Answer
$$\frac{\partial f}{\partial s} \frac{\partial f}{\partial t} =\left(\frac{\partial f}{\partial x}\right)^{2}-\left(\frac{\partial f}{\partial y}\right)^{2}$$
Work Step by Step
Given $$x=s+t,\ \ \ \ \ \ \ y= s-t $$
Since
$$ \frac{\partial x}{\partial s}=1,\ \ \ \frac{\partial y}{\partial s}=1\\
\frac{\partial x}{\partial t}=1,\ \ \ \frac{\partial y}{\partial t}=-1$$
Then
\begin{aligned}
\frac{\partial f}{\partial s} &=\frac{\partial f}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial s}\\
&=\frac{\partial f}{\partial x} \cdot 1+\frac{\partial f}{\partial y} \cdot 1\\
&=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \\
\frac{\partial f}{\partial t} &=\frac{\partial f}{\partial x} \frac{\partial x}{\partial t}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial t}\\
&=\frac{\partial f}{\partial x} \cdot 1+\frac{\partial f}{\partial y} \cdot(-1)\\
&=\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}
\end{aligned}
Hence
\begin{align*}
\frac{\partial f}{\partial s} \frac{\partial f}{\partial t}&=\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right) \left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)\\
&=\left(\frac{\partial f}{\partial x}\right)^{2}-\left(\frac{\partial f}{\partial y}\right)^{2}
\end{align*}
