Let $z=f(u,v)$ where $u=xy$ and $v=\frac{x}{y}$ such that $f$ is second differentiable.
Compute $\tfrac{{\partial}^2z}{\partial x\partial y}$ and $\tfrac{{\partial}^2z}{\partial x^2}$.
My attempt for $\tfrac{{\partial}^2z}{\partial x\partial y}$ is to use the chain rule: \begin{align*} &\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=\dfrac{\partial z}{\partial u}x+\dfrac{\partial z}{\partial v}\dfrac{-x}{y^2} \\ \implies &\dfrac{{\partial}^2z}{\partial x\partial y}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial z}{\partial u}\right)+\dfrac{\partial z}{\partial u}+\dfrac{\partial}{\partial x}\left(\dfrac{\partial z}{\partial v}\right).\dfrac{-x}{y^2}+\dfrac{\partial z}{\partial v}.\dfrac{-1}{y^2}. \end{align*} I don't know how to compute $\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial u}\right)$ and $\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial v}\right)$. I wanna change $x$ to $u, v$ to use the hypothesis $z(u,v)$ is second differetiable.
Could anyone help me? Thanks in advance!
By the chain rule, \begin{align*} \frac{\partial z}{\partial x} &= \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = y \frac{\partial f}{\partial u} + \frac 1y \frac{\partial f}{\partial v}, \text{ and} \\[6pt] \frac{\partial z}{\partial y} &= \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = x \frac{\partial f}{\partial u} - \frac{x}{y^2} \frac{\partial f}{\partial v}. \end{align*} Thus, \begin{align*} \frac{\partial^2 z}{\partial x \partial y} &= \frac{\partial}{\partial x}\left( x \frac{\partial f}{\partial u} - \frac{x}{y^2} \frac{\partial f}{\partial v}\right) \\ &= x \left( y \frac{\partial^2 f}{\partial u^2} + \frac 1y \frac{\partial^2 f}{\partial u \partial v}\right) + \frac{\partial f}{\partial u} - \frac{x}{y^2} \left(y \frac{\partial^2 f}{\partial v \partial u} + \frac 1y \frac{\partial^2 f}{\partial v^2}\right) - \frac{1}{y^2} \frac{\partial f}{\partial v}\\ &= xy \frac{\partial^2 f}{\partial u\partial v} + \frac{\partial f}{\partial u} - \frac{1}{y^2} \frac{\partial f}{\partial v} - \frac{x}{y^3} \frac{\partial^2 f}{\partial v^2}. \end{align*}