Key idea. Use the fixed sum to rewrite everything in terms of the product xy.

1. Since \(x+y=1\), we have \(1-x=y\) and \(1-y=x\). Therefore

   \[1-x^2=y(1+x),\qquad1-y^2=x(1+y).\]

2. Substitute into the expression and cancel positive factors:

   \[\begin{aligned}\left(\frac1{x^2}-1\right)\left(\frac1{y^2}-1\right)&=\frac{(1-x^2)(1-y^2)}{x^2y^2}\\&=\frac{(1+x)(1+y)}{xy}\\&=\frac{2+xy}{xy}=1+\frac2{xy}.\end{aligned}\]

3. Now \((x-y)^2\ge0\) implies \(4xy\le(x+y)^2=1\). Thus \(0<xy\le1/4\), so \(1/(xy)\ge4\).

4. It follows that \(1+2/(xy)\ge1+8=9\). Equality occurs exactly when \(x=y\), which with \(x+y=1\) gives \(x=y=1/2\).

Remember. With a fixed sum, a symmetric expression may simplify to a function of the product.