Solve :

$\dfrac{a}{x} - \dfrac{b}{y} = 0$

$\dfrac{ab^2}{x} + \dfrac{a^2b}{y} = a^2 + b^2$

Let $\dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q$. Substituting in equations, we get :

⇒ ap - bq = 0 ……….(1)

⇒ ab²p + a²bq = a² + b²

⇒ ab²p + a²bq - (a² + b²) = 0 ………(2)

By cross-multiplication method we have :

$\Rightarrow \dfrac{p}{-b \times -(a^2 + b^2) - a^2b \times 0} = \dfrac{q}{0 \times ab^2 - [-(a^2 + b^2)] \times a} = \dfrac{1}{a \times a^2b - ab^2 \times (-b)} \\[1em] \Rightarrow \dfrac{p}{b(a^2 + b^2)} = \dfrac{q}{a(a^2 + b^2)} = \dfrac{1}{a^3b + ab^3} \\[1em] \Rightarrow \dfrac{p}{b(a^2 + b^2)} = \dfrac{1}{a^3b + ab^3} \text{ and } \dfrac{q}{a(a^2 + b^2)} = \dfrac{1}{a^3b + ab^3} \\[1em] \Rightarrow \dfrac{p}{b(a^2 + b^2)} = \dfrac{1}{ab(a^2 + b^2)} \text{ and } \dfrac{q}{a(a^2 + b^2)} = \dfrac{1}{ab(a^2 + b^2)} \\[1em] \Rightarrow p = \dfrac{b(a^2 + b^2)}{ab(a^2 + b^2)} \text{ and } q = \dfrac{a(a^2 + b^2)}{ab(a^2 + b^2)} \\[1em] \Rightarrow p = \dfrac{1}{a} \text{ and } q = \dfrac{1}{b} \\[1em] \Rightarrow \dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{a} \text{ and } \dfrac{1}{y} = \dfrac{1}{b} \\[1em] \Rightarrow x = a \text{ and } y = b.$

Hence, x = a and y = b.