If $\frac{x^2}{a^2+u}+\frac{y^2}{b^2+u}+\frac{z^2}{c^2+u}=1,$ then prove that $(\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}{)}^2+(\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}{)}^2+(\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{z}}{)}^2=2(\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{z}}),$ where $u=u(x,y,z)$ and $a,b,c$ are constants.