#### Answer

$x^2\dfrac{\partial^2 z}{\partial x^2}-y^2(\dfrac{\partial^2 z}{\partial y^2})=(-4uv)\dfrac{\partial^2 z}{\partial u \partial v}+2v\dfrac{\partial z}{\partial v}$

#### Work Step by Step

Given: $z=f(u,v)$
Need to prove that $x^2\dfrac{\partial^2 z}{\partial x^2}-y^2(\dfrac{\partial^2 z}{\partial y^2})=(-4uv)\dfrac{\partial^2 z}{\partial u \partial v}+2v\dfrac{\partial z}{\partial v}$ ....(1)
Now, $\dfrac{\partial z}{\partial x}=\dfrac{\partial f}{\partial u}(y)+\dfrac{\partial f}{\partial v}(\dfrac{-y}{x^2})$
and $\dfrac{\partial z}{\partial y}=\dfrac{\partial f}{\partial u}(x)+\dfrac{\partial f}{\partial v}(\dfrac{1}{x})$
Take second partial derivative.
$\dfrac{\partial^2 z}{\partial x^2}=\dfrac{\partial^2 f}{\partial u^2}(y^2)+(-2y^2/x^2)\dfrac{\partial^2 f}{\partial u \partial v}+\dfrac{\partial f}{\partial v}(2y/x^3)+\dfrac{\partial^2 f}{\partial v^2}(\dfrac{y^2}{x^4})$
and
$\dfrac{\partial^2 z}{\partial y^2}=\dfrac{\partial^2 f}{\partial u^2}(x^2)+(2)\dfrac{\partial^2 f}{\partial u \partial v}+\dfrac{\partial^2 f}{\partial v^2}(x^2)(\dfrac{1}{x^2})$
$y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$
Take left hand side of equation (1).
$x^2\dfrac{\partial^2 z}{\partial x^2}-y^2(\dfrac{\partial^2 z}{\partial y^2})=(-4y^2)\dfrac{\partial^2 f}{\partial u \partial v}+2\dfrac{y}{x}\dfrac{\partial f}{v}$
or, $x^2\dfrac{\partial^2 z}{\partial x^2}-y^2(\dfrac{\partial^2 z}{\partial y^2})=(-4uv)\dfrac{\partial^2 z}{\partial u \partial v}+2v\dfrac{\partial z}{\partial v}$
Hence, it has been proved that $x^2\dfrac{\partial^2 z}{\partial x^2}-y^2(\dfrac{\partial^2 z}{\partial y^2})=(-4uv)\dfrac{\partial^2 z}{\partial u \partial v}+2v\dfrac{\partial z}{\partial v}$