Very interesting multivariable calculus question.

Question

If $\displaystyle z = \frac{f(x-y)}{y}$, show that $\displaystyle z + y \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0$.

Answer 1

$$ \frac{\partial z}{\partial x}= \frac{f'(x-y)}{y}, \\ \frac{\partial z}{\partial y} = \frac{-f'(x-y)y - f(x-y)}{y^2} = -\frac{f'(x-y)}{y} - \frac{f(x-y)}{y^2}. $$

So: $$ z + y\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = \frac{f(x-y)}{y} + y\cdot \frac{f'(x-y)}{y} - f'(x-y) - \frac{f(x-y)}{y} = 0. $$