Answer
We prove that
${v^2}\frac{{\partial g}}{{\partial u}} + {u^2}\frac{{\partial g}}{{\partial v}} = 0$
Work Step by Step
We have $g\left( {u,v} \right) = f\left( {{u^3} - {v^3},{v^3} - {u^3}} \right)$. Write $x = {u^3} - {v^3}$ and $y = {v^3} - {u^3}$.
The partial derivatives of $x$ and $y$ are
$\frac{{\partial x}}{{\partial u}} = 3{u^2}$, ${\ \ \ \ }$ $\frac{{\partial y}}{{\partial u}} = - 3{u^2}$
$\frac{{\partial x}}{{\partial v}} = - 3{v^2}$, ${\ \ \ \ }$ $\frac{{\partial y}}{{\partial v}} = 3{v^2}$
Using the Chain Rule, Eq. (2) and Eq. (3) of Section 15.6, we have
$\frac{{\partial g}}{{\partial u}} = \frac{{\partial f}}{{\partial u}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}$
$\frac{{\partial g}}{{\partial v}} = \frac{{\partial f}}{{\partial v}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial v}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial v}}$
So,
(1) ${\ \ \ \ }$ $\frac{{\partial g}}{{\partial u}} = 3{u^2}{f_x} - 3{u^2}{f_y}$
(2) ${\ \ \ \ }$ $\frac{{\partial g}}{{\partial v}} = - 3{v^2}{f_x} + 3{v^2}{f_y}$
Multiplying equation (1) by ${v^2}$ on both sides gives
(3) ${\ \ \ \ }$ ${v^2}\frac{{\partial g}}{{\partial u}} = 3{u^2}{v^2}{f_x} - 3{u^2}{v^2}{f_y}$
Multiplying equation (2) by ${u^2}$ on both sides gives
(4) ${\ \ \ \ }$ ${u^2}\frac{{\partial g}}{{\partial v}} = - 3{u^2}{v^2}{f_x} + 3{u^2}{v^2}{f_y}$
Adding equation (4) to equation (3) gives
${v^2}\frac{{\partial g}}{{\partial u}} + {u^2}\frac{{\partial g}}{{\partial v}} = 0$
