# The product of two symmetric groups acting on a function

Considering the rational function \small \begin{align*} f&(x_1,x_2,x_3;y_1,y_2,y_3)\\ &=\frac{\left(1-\frac{y_1}{x_1}\right)\left(1-\frac{y_2}{x_1}\right)\left(1-\frac{y_3}{x_1}\right)\left(1-\frac{y_1}{x_2}\right)\left(1-\frac{y_2}{x_2}\right)\left(1-\frac{y_3}{x_2}\right)\left(1-\frac{y_1}{x_3}\right)\left(1-\frac{y_2}{x_3}\right)\left(1-\frac{y_3}{x_3}\right)}{\left(1-\frac{x_2}{x_1}\right)\left(1-\frac{x_3}{x_1}\right)\left(1-\frac{x_3}{x_2}\right)\left(1-\frac{y_2}{y_1}\right)\left(1-\frac{y_3}{y_1}\right)\left(1-\frac{y_3}{y_2}\right)} \\ \end{align*} in $$6$$ variables, I would like to compute the sum $$\sum_{w\in S_3\times S_3}w(f),$$ where the first copy of $$S_3$$ in $$w$$ is permuting the indices of the variables $$x_1, x_2, x_3$$ while the second copy of $$S_3$$ is permuting the indices of the variables $$y_1,y_2,y_3$$.

f[x1_, x2_, x3_, y1_, y2_, y3_] := ((1 - y1/x1) (1 - y2/x1) (1 - y3/x1) (1 - y1/x2) (1 - y2/x2) (1 - y3/x2) (1 - y1/x3) (1 - y2/x3) (1 - y3/x3))/((1 - x2/x1) (1 - x3/x1) (1 - x3/x2) (1 - y2/y1) (1 - y3/y1) (1 - y3/y2))


Sum[