differentiable fn of three variables chain rule

"let f be a differentiable function of three variables, and define w(x,y,z) = f(x-y,y-z,z-x). use the chain rule to show dw/dx + dw/dy + dw/dz = 0". i am confused by this question because, in what i am meant to show, there is no f, i.e. it all comes from the w. so i don't know what part of what i am supposed to differentiate to show this. i also don't know what version of the chain rule i am meant to use. since i have notes on the chain rule that pertain to something entirely different. can someone help? thank you

Re: differentiable fn of three variables chain rule

$\displaystyle \frac{\partial{w}}{dx} = \frac{\partial{f}}{d(x-y)}}\frac{\partial(x-y)}{dx}+ \frac{\partial{f}}{d(y-z)}}\frac{\partial(y-z)}{dx} + \frac{\partial{f}}{d(z-x)}}\frac{\partial(z-x)}{dx} $ = [1, 0, -1]

$\displaystyle \frac{\partial{w}}{dy} = \frac{\partial{f}}{d(x-y)}}\frac{\partial(x-y)}{dy} + \frac{\partial{f}}{d(y-z)}}\frac{\partial(y-z)}{dy} + \frac{\partial{f}}{d(z-x)}}\frac{\partial(z-x)}{dy} $ = [-1, 1, 0]

$\displaystyle \frac{\partial{w}}{dz} = \frac{\partial{f}}{d(x-y)}}\frac{\partial(x-y)}{dz} + \frac{\partial{f}}{d(y-z)}}\frac{\partial(y-z)}{dz} + \frac{\partial{f}}{d(z-x)}}\frac{\partial(z-x)}{dz} $ = [0, -1, 1]

adding [1,0,-1] + [-1,1,0] + [0,-1,1] = [0,0,0]

Re: differentiable fn of three variables chain rule

riiiiight i get it. cool, thanks a lot. it wasn't really presented like the way we've been studying so i wasn't sure how to approach it, but i get the idea now. i could see how the terms 'would' cancel using partials, quite obviously, but didn't know if that was what was required. nice work, thanks