Answer
We prove that
${\left( {\frac{{\partial f}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial f}}{{\partial y}}} \right)^2} = 4u{\left( {\frac{{dg}}{{du}}} \right)^2}$
Work Step by Step
Let $f\left( {x,y} \right) = g\left( u \right)$, where $u = {x^2} + {y^2}$.
The partial derivatives of $u$ are
$\frac{{\partial u}}{{\partial x}} = 2x$, ${\ \ \ \ }$ $\frac{{\partial u}}{{\partial y}} = 2y$
Using the Chain Rule, Eq. (2) and Eq. (3) of Section 15.6, we have
$\frac{{\partial f}}{{\partial x}} = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial x}}$
$\frac{{\partial f}}{{\partial y}} = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial y}}$
Since $f\left( {x,y} \right) = g\left( u \right)$, so $\frac{{\partial f}}{{\partial u}} = \frac{{dg}}{{du}}$.
Thus, we obtain
$\frac{{\partial f}}{{\partial x}} = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial x}} = 2x\frac{{dg}}{{du}}$
$\frac{{\partial f}}{{\partial y}} = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial y}} = 2y\frac{{dg}}{{du}}$
Evaluate ${\left( {\frac{{\partial f}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial f}}{{\partial y}}} \right)^2}$:
${\left( {\frac{{\partial f}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial f}}{{\partial y}}} \right)^2} = 4{x^2}{\left( {\frac{{dg}}{{du}}} \right)^2} + 4{y^2}{\left( {\frac{{dg}}{{du}}} \right)^2} = 4\left( {{x^2} + {y^2}} \right){\left( {\frac{{dg}}{{du}}} \right)^2}$
Since $u = {x^2} + {y^2}$, so ${\left( {\frac{{\partial f}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial f}}{{\partial y}}} \right)^2} = 4u{\left( {\frac{{dg}}{{du}}} \right)^2}$.
