\(\cfrac{x}{y} \propto x+y\)
Thus, \(\cfrac{x}{y} = k_1(x+y)\) [\(k_1\) is a nonzero constant]
\(\cfrac{y}{x} \propto x-y\)
Thus, \(\cfrac{y}{x} = k_2(x-y)\) [\(k_2\) is a nonzero constant]
\(\therefore k_1(x+y)k_2(x-y) = \cfrac{x}{y} \times \cfrac{y}{x}\)
Thus, \(k_1 k_2 (x^2 - y^2) = 1\)
Thus, \(x^2 - y^2 = \cfrac{1}{k_1 k_2} =\) constant (\(\because k_1, k_2\) are constants) (Proved)