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Suppose $z = \ln\Big(\frac{x^3+y^3}{x-y}\Big)$. Show that $x\frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=2.$ Any help much appreciated.

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Calculate the partial derivatives using the chain and quotient rules: $$ \frac{\partial z}{\partial x} = \frac{3x^2(x-y)-(x^3+y^3)}{(x-y)^2} \cdot \frac{(x-y)}{x^3+y^3} $$ $$ \frac{\partial z}{\partial y} = \frac{3y^2(x-y)-(x^3+y^3)(-1)}{(x-y)^2} \cdot \frac{(x-y)}{x^3+y^3} $$ Now all you have to do is to simplify $x\frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}$ with the above expressions inserted and you will get the desired result of 2.

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I might be able to make this simpler. Start by rewriting $z$ as

$$z=\ln(x^3+y^3)-\ln(x-y)$$ $$\frac{\partial z}{\partial x}=\frac{3x^2}{x^3+y^3}-\frac1{x-y}$$ $$\frac{\partial z}{\partial y}=\frac{3y^2}{x^3+y^3}+\frac1{x-y}$$ Now the calculation should be a lot easier.

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