显然，当 $x<0$ 时，$f(x)<0$，当 $x>0$ 时，$f(x)>0$．故只需考察 $x>0$．$f\left(x \right)=\dfrac{x}{\sqrt{a-{{x}^{2}}}-\sqrt{b-{{x}^{2}}}}=\dfrac{x\left(\sqrt{a-{{x}^{2}}}+\sqrt{b-{{x}^{2}}} \right)}{a-b}=\dfrac{1}{a-b}\sqrt{{{x}^{2}}\left(a+b \right)-2{{x}^{4}}+2{{x}^{2}}\sqrt{\left( a-{{x}^{2}} \right)\left(b-{{x}^{2}} \right)}}\\=\dfrac{1}{a-b}\sqrt{ab-{{\left[ {{x}^{2}}-\sqrt{\left(a-{{x}^{2}} \right)\left( b-{{x}^{2}} \right)} \right]}^{2}}}\leqslant \dfrac{\sqrt{ab}}{a-b}.$
当 $x^2=\sqrt{(a-x^2)(b-x^2)}$，即 $x=\sqrt{\dfrac{ab}{a+b}}$ 时取得等号．