Let $z=xy+f(\frac yx) $. Show $z$ satisfies the partial differential eq. $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2xy$

Question

Suppose that $f: R\rightarrow R$ is differentiable and $z=xy+f(\frac yx) $, $(x,y)\in R^2, x\neq 0.$ Show $z$ satisfies the partial differential equation:

$$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2xy$$

From my understanding I differentiate $z$ with respect to some variable, but in this case I'm not sure what it is?

Answer 1

$\frac{\partial z}{\partial x} = y + f'(\frac{y}{x}).\frac{-y}{x^2}$ (treating y as constant)

$ \frac{\partial z}{\partial y} = x + f'(\frac{y}{x}).\frac{1}{x}$ (treating x as constant)

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$x\frac{\partial z}{\partial x} = xy + f'(\frac{y}{x}).\frac{-y}{x}$

$ y\frac{\partial z}{\partial y} = xy + f'(\frac{y}{x}).\frac{y}{x}$

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$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 2xy$

Answer 2

We get $$xy-\frac{y}{x}f_x(\frac{y}{x})+xy+\frac{y}{x}f_y(\frac{y}{x})$$ and if $$f_x=f_y$$ we get $$2xy$$