Now,
$
\dfrac{x^2}{1+u}+\dfrac{y^2}{2+u}+\dfrac{z^2}{3+u}=1
$
Differentiating given equation partially w.r.t. x,
$
\therefore
\dfrac{2x}{2+u} -
\dfrac{x^2}{(1+u)^2}
\dfrac{\partial u}{\partial x} -
\dfrac{y^2}{(2+u)^2}
\dfrac{\partial u}{\partial x} -
\dfrac{z^2}{(3+u)^2}
\dfrac{\partial u}{\partial x}
\\ \; \\ \; \\
\therefore
\dfrac{2x}{1+u} \;=\;
\dfrac{\partial u}{\partial x}
\bigg[
\dfrac{x^2}{(1+u)^2}+
\dfrac{y^2}{(2+u)^2}+
\dfrac{z^2}{(3+u)^2}
\bigg]
\\ \; \\ \; \\
\therefore
\dfrac{2x}{1+u} \;=\; p . \dfrac{\partial u}{\partial x}
\; where \;
\; p\;=\;
\dfrac{x^2}{(1+u)^2}+
\dfrac{y^2}{(2+u)^2}+
\dfrac{z^2}{(3+u)^2} \; \; \; \ldots (i)
\\ \; \\ \; \\
\therefore
\dfrac{\partial u}{\partial x} \;=\;
\dfrac{2x}{(1+u) p}
$
Similarly, we can find out
$
\therefore
\dfrac{\partial u}{\partial y} \;=\;
\dfrac{2y}{(1+u) p} \; and \;
\dfrac{\partial u}{\partial z} \;=\;
\dfrac{2z}{(1+u) p}
\\ \; \\ \; \\
\bigg(\dfrac{\partial u}{\partial x} \bigg)^2 +
\bigg(\dfrac{\partial u}{\partial y} \bigg)^2 +
\bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\;
\dfrac{4x^2}{(1+u)^2 . p} +
\dfrac{4y^2}{(2+u)^2 . p} +
\dfrac{4z^2}{(3+u)^2 . p}
\\ \; \\ \; \\
=
\dfrac{4}{p^2}
\bigg[
\dfrac{x^2}{(1+u)^2}+
\dfrac{y^2}{(2+u)^2}+
\dfrac{z^2}{(3+u)^2}
\bigg]
\\ \; \\ \; \\
=
\dfrac{4}{p^2} . p
\\ \; \\ \; \\
\bigg(\dfrac{\partial u}{\partial x} \bigg)^2 +
\bigg(\dfrac{\partial u}{\partial y} \bigg)^2 +
\bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\;
\dfrac{4}{p}
\; \; \; \; \; \ldots (ii)
$
Now,
$
\therefore
x\dfrac{\partial u}{\partial x} +
y\dfrac{\partial u}{\partial y} +
z \dfrac{\partial u}{\partial z}
\;=\;
\dfrac{2x^2}{(1+u). p} +
\dfrac{2y^2}{(2+u). p} +
\dfrac{2z^2}{(3+u). p}
\\ \; \\ \; \\
\therefore
x\dfrac{\partial u}{\partial x} +
y\dfrac{\partial u}{\partial y} +
z \dfrac{\partial u}{\partial z}
\;=\;
\dfrac{2}{p} \Bigg[
\dfrac{x^2}{(1+u)} +
\dfrac{y^2}{(2+u)} +
\dfrac{z^2}{(3+u)}
\Bigg]
\\ \; \\ \; \\
= \dfrac{2}{p} (1)
\\ \; \\ \; \\
\therefore
x\dfrac{\partial u}{\partial x} +
y\dfrac{\partial u}{\partial y} +
z \dfrac{\partial u}{\partial z}
\;=\; \dfrac{2}{p}
$
From (ii) and (iii), we can prove that
$
\bigg(\dfrac{\partial u}{\partial x} \bigg)^2 +
\bigg(\dfrac{\partial u}{\partial y} \bigg)^2 +
\bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\;
2 \bigg(
x\dfrac{\partial u}{\partial x} +
y\dfrac{\partial u}{\partial y} +
z \dfrac{\partial u}{\partial z}
\bigg)
$