The key to understanding differentials, at least in practical use, is as a most general form of differentiation -- one done with respect to no variables, or else to a parameter which we must assume every variable involved is a function of. This is the key to the statement $y=x \implies dy = dx$, and its (semantically, deeper) equivalent statement $\frac{dy}{dx} = 1$ .
So, $xyz = c \implies d(xyz) = d(c)$ , and we apply usual rules of differentiation, as though each variable invokes a chain rule: $ d(xyz) = xy~dz + x~dy~z + dx~yz ~~,~~d(c) = 0 \text{ as c is constant.} $ So solve for $dz$, and you should arrive at the same solution.