If  $f(x+y,x-y)=xy$, then the arithmetic mean of  $f(x,y)$ and  $f(y,-x)$ is

$f(x+y,x-y)=xy$

$x+y=a$

$x-y=b$

$x=\frac{a+b}{2}$

$y=\frac{a-b}{2}$

Replacing with $x$ with  $\frac{x+y}{2}$ and  $y$ with  $\frac{x-y}{2}$

$f(x,y)=\left(\frac{x+y}{2}\right)\left(\frac{x-y}{2}\right)$

$\left.\begin{array}{r}\begin{array}{l}f(x,y)=\frac{x^2-y^2}{4}\\ f(y,-x)=\frac{y^2-x^2}{4}\end{array}\\ \end{array}\right\} \text{AM}=\frac{f(x,y)+f(y,-x)}{2}=0$