The equation $x^2\frac{\partial^2 z}{\partial x^2}-2xy\frac{\partial^2 z}{\partial x\partial y}+y^2\frac{\partial^2 z}{\partial y^2}=0$ can be transformed by substituting variables
$\begin{cases}u=xy\\v=\frac{1}{y}\end{cases}$
I have attached my solution in the images below showing how I arrived to $$2v^2\frac{\partial^2 z}{\partial v^2}+v\frac{\partial z}{\partial v}-u\frac{\partial z}{\partial u}=0$$
I expected both second partials of u and v to vanish and have been stuck wondering if this could have a simpler form without second partials, or something with mixed second partials. Any sort of pointers on this would be helpful
Your first mistake is in the beginning.
$$\frac{\partial v}{\partial x} = \color{red}{0} \ne \frac{1}{u}$$
Second, your chain rule application isn't quite correct. See this answer for my full derivation.
As a reminder, note that since
$$ \frac{\partial }{\partial x}(z) = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x} $$
we have
$$ \frac{\partial }{\partial x}\left(\frac{\partial z}{\partial u}\right) = \frac{\partial}{\partial u}\left( \frac{\partial z}{\partial u}\right)\frac{\partial u}{\partial x} + \frac{\partial }{\partial v}\left( \frac{\partial z}{\partial u}\right)\frac{\partial v}{\partial x} = \frac{\partial^2 z}{\partial u^2} \frac{\partial u}{\partial x} + \color{blue}{\frac{\partial^2 z}{\partial u\partial v} \frac{\partial v}{\partial x}} $$
As a result
$$ \frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial u^2} \left(\frac{\partial u}{\partial x}\right)^2 + \color{blue}{2\frac{\partial^2 z}{\partial u \partial v} \frac{\partial u}{\partial x} \frac{\partial v}{\partial x}} + \frac{\partial^2 z}{\partial v^2}\left( \frac{\partial v}{\partial x}\right)^2 + \frac{\partial z}{\partial u} \frac{\partial^2 u}{\partial x^2} + \frac{\partial z}{\partial v} \frac{\partial^2 v}{\partial x^2} $$
See if you can find the remaining derivatives in the same way....
The correct derivatives are
\begin{align} \frac{\partial^2 z}{\partial x^2} &= y^2 \frac{\partial^2 z}{\partial u^2} \\ \frac{\partial^2 z}{\partial x \partial y} &= xy \frac{\partial^2 z}{\partial u^2} - \frac{1}{y} \frac{\partial^2 z}{\partial u \partial v} + \frac{\partial z}{\partial u} \\ \frac{\partial^2 z}{\partial y^2} &= x^2 \frac{\partial^2 z}{\partial u^2} - \frac{2x}{y^2} \frac{\partial^2 z}{\partial u\partial v} + \frac{1}{y^4} \frac{\partial^2 z}{\partial v^2} + \frac{2}{y^3} \frac{\partial z}{\partial v} \end{align}
And the final equation turns out to be
$$ v^2 \frac{\partial^2 z}{\partial v^2} + 2v \frac{\partial z}{\partial v} - 2u\frac{\partial z}{\partial u} = 0 $$