Let $z=f(x,y)$ and $x=e^{u}cos\, v\; $ and $y=e^{u}sin\,v\;$and $f$ have continuous second-order partial derivatives.
i. Show $$\frac{\partial^{2}z}{\partial{u}^2}=x^2\frac{\partial^{2}f}{\partial{x}^2}+2xy\frac{\partial^{2}f}{\partial{x}\partial{y}}+y^2\frac{\partial^{2}f}{\partial{x}^2}+x\frac{\partial{f}}{\partial{x}}+y\frac{\partial{f}}{\partial{y}}$$
I'm able to show RHS excluding the mixed partial derivative. Can't figure out where that comes from. Please help.
$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$
$\frac{\partial^2z}{\partial u^2}=\frac{\partial^2 z}{\partial x^2}(\frac{\partial x}{\partial u})^2+2\frac{\partial^2z}{\partial x\partial y}\frac{\partial x}{\partial u}\frac{\partial y}{\partial u}+\frac{\partial^2 z}{\partial y^2}(\frac{\partial y}{\partial u})^2+\frac{\partial z}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial z}{\partial y}\frac{\partial^2 y}{\partial u^2}$
Using the definition of $x$ and $y$, the result follows.