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If z = f(x,y) , x=(eu + e-v), y=(e-u - ev) \[ \dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v} = x\dfrac{\partial z}{\partial x}-y\dfrac{\partial z}{\partial y}\]
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$\dfrac {\partial z}{\partial u} = \dfrac {\partial z}{\partial x} \dfrac {\partial x}{\partial u}+ \dfrac {\partial z}{\partial y}\dfrac {\partial y}{\partial u} = \dfrac {\partial z}{\partial x} e^{u} - \dfrac {\partial z}{\partial y}e^{-u}$ $\dfrac {\partial z}{\partial v} = \dfrac {\partial z}{\partial x} \dfrac {\partial x}{\partial v} + \dfrac {\partial z}{\partial y}\dfrac {\partial y}{\partial v} = - \dfrac {\partial z}{\partial x} e^{-v} - \dfrac {\partial z}{\partial y} e^v$ By subtraction, $\dfrac {\partial z}{\partial u} - \dfrac {\partial z}{\partial v} = (e^u + e^{-v}) \dfrac {\partial z}{\partial x } (e^{-u} -e^v) \dfrac {\partial z}{\partial y} = x \dfrac {\partial z}{\partial x}- y \dfrac {\partial z}{\partial y}$

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