One of the side effects of learning on your own is to seriously doubt if the book has an error or your reasoning is incorrect. My book on multivariable calculus (Rogawski 3rd Edition of Calculus Multivariable) has a problem at the end of Chapter 15:
Let $g(u, v) = f(u^3-v^3, v^3 - u^3)$. Prove that, $$ v^2 \frac{\partial g}{\partial u} - u^2 \frac{\partial g}{\partial v} = 0 $$
I proceed as following: Consider the functions $x(u, v) = u^3 - v^3$ and $y(u,v) = v^3 - u^3$. I can now say $g(u, v) = f(x, y)$. And therefore, $$ \frac{\partial g}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u} \\ \frac{\partial g}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} $$
Now, $$ \frac{\partial x}{\partial u} = 3u^2 \\ \frac{\partial x}{\partial v} = -3v^2 \\ \frac{\partial y}{\partial u} = -3u^2 \\ \frac{\partial y}{\partial v} = 3v^2 $$ Therefore,
\begin{align} & \frac{\partial g}{\partial u} = 3u^2\frac{\partial f}{\partial x} - 3u^2 \frac{\partial f}{\partial y} \\ \implies & \frac{1}{3u^2} \frac{\partial g}{\partial u} = \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y} \\ & \frac{\partial g}{\partial v} = -3v^2\frac{\partial f}{\partial x} + 3v^2\frac{\partial f}{\partial y} \\ \implies & \frac{1}{3v^2} \frac{\partial g}{\partial v} = -\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \end{align}
Adding the two up I get $\displaystyle v^2\frac{\partial g}{\partial u} + u^2\frac{\partial g}{\partial v} = 0$ which is different from what I am asked to prove by a minus sign.
Am I wrong or is there a typing error in the book?