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Partial derivatives chain rule (calculus II) I was doing some exercises and I came upon this one. I have been thinking about for quite some time, but I'm still not sure about the answer. Any help would be gladly appreciated. !Problem Link This is what I have tried: !Attempted Solution Is this correct?

Rewrite $$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}=2x \frac{\partial z}{\partial u}+2y \frac{\partial z}{\partial v}.$$ (Similarly for $\frac{\partial z}{\partial y}$.)

This results in $$2(x^2-y^2)\frac{\partial z}{\partial u}=0.$$

So, the transformed differential equation is $$\frac{\partial \, z(u,v)}{\partial u}=0.$$

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