Prove that if $f$ is an arbtirary $C^1$ function of two variables, then: $$z=\frac{1}{3}x_1x_2x_3-f(\frac{x_2}{x_1},\frac{x_3}{x_1}), x_1\ne 0$$
satisfies the equation $$x_1 \frac{\partial z}{\partial x_1} + x_2 \frac{\partial z}{\partial x_2} + x_3 \frac{\partial z}{\partial x_3} = x_1 x_2 x_3$$
Working out: I know that the answer comes from the first term in the equation of z - I would have: $$\frac{\partial z}{\partial x_1}= \frac{1}{3}x_2 x_3 - f'$$ $$\frac{\partial z}{\partial x_2}= \frac{1}{3}x_1 x_3 - f'$$ $$\frac{\partial z}{\partial x_3}= \frac{1}{3}x_1 x_2 - f'$$ and therefore $$x_1 \frac{\partial z}{\partial x_1} + x_2 \frac{\partial z}{\partial x_2} + x_3 \frac{\partial z}{\partial x_3} = 3 \times \frac{1}{3}x_1 x_2 x_3- x_1 f' - x_2 f' - x_3f'$$
I just don't know about how to differentiate the second part of the z term. I can't understand how f can be a function of 2 variables - wouldn't it have $x_1, x_2, x_3$ terms in it? I have no idea how I would express the partial derivative of f (which I just called f' above) and how to fit it into the equation. Thanks for any help!
For $t>0$ let $g(t):=z(tx_1, tx_2, tx_3)=\frac13 t^3 x_1 x_2 x_3-f(\frac{x_2}{x_1},\frac{x_3}{x_1})$. Then $g'(t)=t^2 x_1 x_2 x_3$ on the one hand. On the other hand $g'(t)=\frac{d}{dt}(t x_1)\frac{\partial }{\partial x_1}z(tx_1,tx_2,tx_3)+\frac{d}{dt}(t x_2)\frac{\partial }{\partial x_2}z(tx_1,tx_2,tx_3)+\frac{d}{dt}(t x_3)\frac{\partial }{\partial x_3}z(tx_1,tx_2,tx_3)$. Observe that $\frac{d}{dt}(t x_i)=x_i$ for $i=1,2,3$. Finally put $t=1$.