Using Monge's method $(r-s)x = (t-s)y$ 
Solve using Monge's method
$$(r-s)x=(t-s)y$$
Where

*

*$r={\partial ^2z\over \partial x^2}$,

*$s={\partial ^2z\over \partial x\partial y}$ and

*$t={\partial ^2z\over \partial y^2}$.


My attempt:
I found out two distinct intermediate integrals and solved them for $p$ and $q$, but I got stuck after this step:
$$
dz=(ydx-xdy).\phi\left({x\over y}\right)+(dx+dy)\cdot\big(z+\psi(x+y)\big),
$$
where $\phi$ and $\psi$ are arbitrary functions. Here $z$ is creating some problems.
My attempt was to transform the equation in an exact differential, something like the following form
$$
d\big(z\cdot f(x,y)\big)=\phi'\left({x\over y}\right)d\left({x\over y}\right)+\psi'(x+y)d(x+y)$$
 A: Using Monge's method, the equation reads
$$x r-yt+(y-x)s=0\tag{1}$$
where $r=\partial_{xx}z$, $t=\partial_{yy}z$ and $s=\partial_{xy}z$ (I denote partial derivative $\frac{\partial^2 z}{\partial x^2}$ as $\partial_{xx}z$ etc). Set $p=\partial_x z$ and $q=\partial_y z$ so that
$$dz=pdx+q dy.$$
From $dp=rdx+sdy$ and $dq=sdx+tdy$ we get
$$r=\frac{dp-sdy}{dx},\qquad t=\frac{dq-sdx}{dy}.$$
Plugging this into the PDE we get:
$$x\left(\frac{dp-sdy}{dx}\right)-y\left(\frac{dq-sdx}{dy}\right)+(y-x)s=0.$$
Multiplying by $dxdy$ and rearranging terms we get:
$$xdpdy-ydqdx-s[(x-y)dxdy+xdy^2-ydx^2]=0.$$
The two Monge auxiliary equations are therefore:
$$xdpdy-ydqdx=0,\tag{2a}$$
$$(x-y)dxdy+xdy^2-ydx^2=0\tag{2b}.$$
Next we write $(2b)$ as
$$(xdy-ydx)(dx+dy)=0\tag{3}.$$
Now the Monge method considers the case where the first factor cancels: $xdy-ydx=0$ which means $x/y=c_1$, and since $ydx=xdy$ equation $(2a)$ becomes merely $dp=dq$ so that $p-q=c_2$. This means that $p-q=f(\frac xy)$. When the second factor cancels, $dx=-dy$ thus
$x+y=c_3$ and
$$dz=pdx+qdy=(p-q)dx=h(\frac x y)dx$$
for arbitrary $f$. Integrating gives the general solution to $(1)$ as
$$z(x,y)=y H(\frac x y)+M(x+y)$$
where $H$ and $M$ are arbitrary.

One more modern way to integrate your PDE which is :
$$x(\partial_{xx}z-\partial_{xy}z)=y(\partial_{yy}z-\partial_{xy}z)$$
is to use the d'Alembert change of variables as in the wave equation. Set
$$\xi=x+y\qquad \eta=x-y$$
and plug $z(x,y)=v(\xi,\eta)=v(x+y,x-y)$ into the equation. By the chain rule, you eventually get a PDE on $v$ which is
$$\xi\partial_{\xi\eta}v+\eta\partial_{\eta\eta}v=0.$$
This one is easy to integrate since $w=\partial_\eta v$ should verify:
$$\xi\partial_\xi w+\eta\partial_\eta w=0,$$
whose integral is $w(\xi,\eta)=f(\frac\eta\xi)$ for an arbitrary $f$.
Integrating in $\eta$ gives
$$v(\xi,\eta)=\xi F(\frac\eta\xi)+G(\xi)$$ for arbitrary $F$ and $G$. Therefore the general integral of your PDE is given by:
$$z(x,y)=(x+y)F(\frac{x-y}{x+y})+G(x+y)$$
for arbitrary functions $F$ and $G$. Note that since
$$(x+y)F(\frac{x-y}{x+y})=y(1+\frac x y)F(\frac{x/y-1}{x/y+1})$$
this solution could also be written as
$$z(x,y)=yM(\frac x y)+G(x+y)$$
so that we recover the same solution as by Monge's method.
