We have received a very clearly explained solution to this, but unfortunately whoever sent it did not include their name. If it was you, let us know!

Let ab=x and cd=y (where ab means a as the tens digit, and b as the ones digit, not a times b).

${\displaystyle \frac{{\mathrm{abcd}}^2-{\mathrm{cdab}}^2}{{\mathrm{ab}}^2-{\mathrm{cd}}^2}=\frac{(100x+y{)}^2-(100y+x{)}^2}{x^2-y^2}}$

${\displaystyle =\frac{(10000x^2+200\mathrm{xy}+y^2)-(10000y^2+200\mathrm{xy}+x^2)}{x^2-y^2}}$

${\displaystyle =\frac{9999x^2-9999y^2}{x^2-y^2}}$

${\displaystyle =\frac{9999(x^2-y^2)}{x^2-y^2}}$

${\displaystyle =9999}$

(since x>y we are not dividing by zero)