Let the fourth proportional of \((x^2 - y^2), (x^2y - xy^2), (x + y)\) be \(p\). \[ \therefore \frac{(x^2 - y^2)}{(x^2y - xy^2)} = \frac{(x + y)}{p} \Rightarrow p(x^2 - y^2) = (x + y)(x^2y - xy^2) \Rightarrow p = \frac{(x + y)(x^2y - xy^2)}{(x^2 - y^2)} \Rightarrow p = \frac{(x + y) \cdot xy(x - y)}{(x + y)(x - y)} \Rightarrow p = \frac{\cancel{(x + y)} \cdot xy \cancel{(x - y)}}{\cancel{(x + y)} \cancel{(x - y)}} \Rightarrow p = xy \] ∴ The fourth proportional of \((x^2 - y^2), (x^2y - xy^2), (x + y)\) is \(xy\).