Hi.

So, given $\displaystyle f(x,y,z) = 0 $ and $\displaystyle g(x,y,z) = 0$

Show that $\displaystyle \frac{dx}{\frac{\delta (f,g)}{\delta (y,z)}} = \frac{dy}{\frac{\delta (f,g)}{\delta (x,z)}} = \frac{dz}{\frac{\delta (f,g)}{\delta (x,y)}}$

I think it makes sense (I can "see" why the equality holds) but can't think of a way to derive it. To begin, I wrote $\displaystyle f(x,y,z) = 0 $ as $\displaystyle f(x_f(y,z),y_f(x,z),z_f(x,y)) = 0 $, assuming each argument was an implicit function of the others. Similarly, $\displaystyle g(x_g(y,z),y_g(x,z),z_g(x,y)) = 0 $

I feel like some version of the chain rule would be useful ( does the cross product of the gradients have some relationship to differential??), and that the above equality could be easier to derive if written as $\displaystyle \frac{dy}{dx} = \frac{\frac{\delta (f,g)}{\delta (x,z)}}{\frac{\delta (f,g)}{\delta (y,z)}} = .... $. I seem to have trouble taking the gradient,or any partial derivatives of implicit functions when using the chain rule... ie, is $\displaystyle \frac{\delta f}{\delta x_f(y,z)}\frac{\delta x_f(y,z)}{\delta (y,z)} $ the correct derivative of f w.r.t x?

Any help appreciated, thanks.