Prove that $x=u^2v$ and $y=uv^2$ satisfy a certain PDE Given that $z$ is a function of $x$ and $y$ and that $x=u^2v$ and $y=uv^2$, i tried to prove that
$$2x^2\frac{\partial^2z}{\partial x^2}+5xy\frac{\partial^2z}{\partial x\partial y}+2y^2\frac{\partial^2z}{\partial y^2}=uv\frac{\partial^2z}{\partial u\partial v}-\frac{2}{3}\left(u\frac{\partial z}{\partial u}+v\frac{\partial z}{\partial v}\right).$$
Here's how I go about:
Given $x=u^2v$ and $y=uv^2$, we have,
$$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}=2uv\frac{\partial z}{\partial x}+v^2\frac{\partial z}{\partial y}$$
and
$$\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}=u^2\frac{\partial z}{\partial x}+2uv\frac{\partial z}{\partial y}.$$
Solving these two equations for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$, we get,
$$\frac{\partial z}{\partial x}=\frac{2}{3uv}\frac{\partial z}{\partial u}-\frac{1}{3u^2}\frac{\partial z}{\partial v}\quad\implies\quad \frac{\partial}{\partial x}=\frac{2}{3uv}\frac{\partial}{\partial u}-\frac{1}{3u^2}\frac{\partial}{\partial v}$$
and
$$\frac{\partial z}{\partial y}=-\frac{1}{3v^2}\frac{\partial z}{\partial u}+\frac{2}{3uv}\frac{\partial z}{\partial v}\quad\implies\quad \frac{\partial}{\partial y}=-\frac{1}{3v^2}\frac{\partial}{\partial u}+\frac{2}{3uv}\frac{\partial}{\partial v}$$
Calculating $\frac{\partial^2z}{\partial x^2}$, $\frac{\partial^2z}{\partial x\partial y}$ and $\frac{\partial^2z}{\partial y^2}$, I got
$$2x^2\frac{\partial^2z}{\partial x^2}+5xy\frac{\partial^2z}{\partial x\partial y}+2y^2\frac{\partial^2z}{\partial y^2}=uv\frac{\partial^2z}{\partial u\partial v}+v^2\frac{\partial^2z}{\partial v^2}.$$
I am at a loss as how to obtain
$$v^2\frac{\partial^2z}{\partial v^2}=-\frac{2}{3}\left(u\frac{\partial z}{\partial u}+v\frac{\partial z}{\partial v}\right).$$
 A: Lets see: $x=u^2v$ and $y=uv^2$, so ... as you began:
$$\small\begin{align}\dfrac{\partial z}{\partial u}&=\dfrac{\partial x}{\partial u}\cdotp\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial u}\cdotp\dfrac{\partial z}{\partial y}\\[1ex]&=2uv\cdotp\dfrac{\partial z}{\partial x}+v^2\cdotp\dfrac{\partial z}{\partial y}&\therefore~&u\cdotp\dfrac{\partial z}{\partial u}=2x\cdotp\dfrac{\partial z}{\partial x}+y\cdotp\dfrac{\partial z}{\partial y}\\[2ex]\dfrac{\partial z}{\partial v}&=\dfrac{\partial x}{\partial v}\cdotp\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial v}\cdotp\dfrac{\partial z}{\partial y}\\[1ex]&=u^2\cdotp\dfrac{\partial z}{\partial x}+2uv\cdotp\dfrac{\partial z}{\partial y}&\therefore~&v\cdotp\dfrac{\partial z}{\partial u}=x\cdotp\dfrac{\partial z}{\partial x}+2y\cdotp\dfrac{\partial z}{\partial y}\end{align}$$
So you immediately have that $\small\dfrac{2}{3}\left(u\cdotp\dfrac{\partial z}{\partial u}+v\cdotp\dfrac{\partial z}{\partial v}\right)=2x\cdotp\dfrac{\partial z}{\partial x}+2y\cdotp\dfrac{\partial z}{\partial y}$

Then to obtain the rest of the RHS just take another derivation, applying the product rule then the chain rule ($\times2$):
$$\small\begin{align}\dfrac{\partial^2 z}{\partial u~\partial v}&=\dfrac{\partial }{\partial u}\left(u^2\cdotp\dfrac{\partial z}{\partial x}+2uv\cdotp\dfrac{\partial z}{\partial y}\right)\\[1ex]&=2u\cdotp\dfrac{\partial z}{\partial x}+u^2\cdotp\dfrac{\partial~~}{\partial u}\dfrac{\partial z}{\partial x}+2v\cdotp\dfrac{\partial z}{\partial y}+2uv\cdotp\dfrac{\partial~~}{\partial u}\dfrac{\partial z}{\partial y}\\[1ex]&=2u\cdotp\dfrac{\partial z}{\partial x}+u^2\left(\dfrac{\partial x}{\partial u}\cdotp\dfrac{\partial^2 z}{\partial x~^2}+\dfrac{\partial y}{\partial u}\cdotp\dfrac{\partial^2 z}{\partial x~\partial y}\right)+2v\cdotp\dfrac{\partial z}{\partial y}+2uv\left(\dfrac{\partial x}{\partial u}\cdotp\dfrac{\partial^2 z}{\partial x~\partial y}+\dfrac{\partial y}{\partial u}\cdotp\dfrac{\partial^2 z}{\partial y~^2}\right)\\[1ex]&=2u\cdotp\dfrac{\partial z}{\partial x}+u^2\left(2uv\cdotp\dfrac{\partial^2 z}{\partial x~^2}+v^2\cdotp\dfrac{\partial^2 z}{\partial x~\partial y}\right)+2v\cdotp\dfrac{\partial z}{\partial y}+2uv\left(2uv\cdotp\dfrac{\partial^2 z}{\partial x~\partial y}+v^2\cdotp\dfrac{\partial^2 z}{\partial y~^2}\right)\\[1ex]&=2 u\cdotp\dfrac{\partial z}{\partial x}+2u^3v\cdotp\dfrac{\partial^2 z}{\partial x~^2}+5u^2v^2\cdotp\dfrac{\partial^2 z}{\partial x~\partial y}+2uv^3\cdotp\dfrac{\partial^2 z}{\partial y~^2}+2v\cdotp\dfrac{\partial z}{\partial y}\\[2ex]uv\cdotp\dfrac{\partial^2 z}{\partial u~\partial v}&=2 x\cdotp\dfrac{\partial z}{\partial x}+2x^2\cdotp\dfrac{\partial^2 z}{\partial x~^2}+5xy\cdotp\dfrac{\partial^2 z}{\partial x~\partial y}+2y^2\cdotp\dfrac{\partial^2 z}{\partial y~^2}+2y\cdotp\dfrac{\partial z}{\partial y}\end{align}$$
And the rest is algebra.
