Đặt: $\dfrac{x}{a}=s,\dfrac{y}{b}=t,\dfrac{z}{c}=u$.
Ta có: $s+t+u=1$
Lại có: $\dfrac{1}{s}+\dfrac{1}{t}+\dfrac{1}{u}=0 \Leftrightarrow \dfrac{st+tu+su}{stu}=0 \Leftrightarrow st+tu+su=0$
Từ đó suy ra:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=s^2+t^2+u^2=(s+t+u)^2-2(st+tu+su)=1$.