if
$$\large w=f\big(\frac{1}{x} - \frac{1}{y} , \frac{1}{x} - \frac{1}{z}\big)$$
what is the following expression equal to:
$$\large x^2\frac{\partial{w}}{\partial{x}}+y^2\frac{\partial{w}}{\partial{y}}+z^2\frac{\partial{w}}{\partial{z}}=?$$
this is what I've done so far:
$$\large u(x,y)=\frac{1}{x} - \frac{1}{y}\\ \large v(x,z)=\frac{1}{x} - \frac{1}{z}$$
and now we can do chain rule for $f(u,v)$:
$$\large \frac{\partial{w}}{\partial{x}}=\frac{\partial{w}}{\partial{u}}.\frac{\partial{u}}{\partial{x}} + \frac{\partial{w}}{\partial{v}}.\frac{\partial{v}}{\partial{x}}\\\large \frac{\partial{w}}{\partial{y}}=\frac{\partial{w}}{\partial{u}}.\frac{\partial{u}}{\partial{y}} + \frac{\partial{w}}{\partial{v}}.\frac{\partial{v}}{\partial{y}}\\\large \frac{\partial{w}}{\partial{z}}=\frac{\partial{w}}{\partial{u}}.\frac{\partial{u}}{\partial{z}} + \frac{\partial{w}}{\partial{v}}.\frac{\partial{v}}{\partial{z}}$$
and by substituting these in the above expression I got this:
$$\large x^2.\frac{-1}{x^2}(\frac{\partial{w}}{\partial{u}}+\frac{\partial{w}}{\partial{v}})+y^2.\frac{1}{y^2}(\frac{\partial{w}}{\partial{u}}+\frac{\partial{w}}{\partial{v}})+z^2.\frac{1}{z^2}(\frac{\partial{w}}{\partial{v}})=\frac{\partial{w}}{\partial{v}}$$
now I want to know if I can take this any further or should I stop at this and if am I allowed to make the substitution in the first place
Thank you in advance
As Anne Bauval mentioned the $\frac{\partial v}{\partial y}$ will be equal to zero and therefore the final answer would be $0$ and not $\frac{\partial w}{\partial v}$