If $\dfrac{a}{b} = \dfrac{c}{d}$, then $\dfrac{a^2+b^2}{a^2-b^2}$ also equals—

Substituting $a=ck,\ b=dk$ gives $\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{k^2(c^2+d^2)}{k^2(c^2-d^2)}=\dfrac{c^2+d^2}{c^2-d^2}$. The factor $k^2$ cancels, leaving the expression in terms of $c$ and $d$.