Simultaneous P.D.E. problem 
Solve the following differential equation:
  $$ \frac{dx}{y^2(x-y)} = \frac{dy}{x^2(x-y)} = \frac{dz}{z(x^2 + y^2)} $$

I have so far got
$x^2dx = y^2dy $ which implies that $(x-y)(x^2 +xy + y^2) = a$ ...(1)
Then $$\frac{dx+ dy}{(x-y)} = \frac{dz}{z}$$
Using the equation (1), I get 
$\frac{1}{a} (x^2+xy+y^2)(dx+dy) = \frac{dz}{z}$
How do I proceed from this point? I could have solved this question easily if the third fraction in the given P.D.E. were
$\frac{dz}{z(x^2 - y^2)}$ but I need help for this problem. Please help me out.
 A: $$ \frac{dx}{y^2(x-y)} = \frac{dy}{x^2(x-y)} = \frac{dz}{z(x^2 + y^2)} $$
This system of ODEs might come from solving the PDE :
$$y^2(x-y)\frac{\partial z}{\partial x}+x^2(x-y)\frac{\partial z}{\partial y}=(x^2+y^2)z$$
A first characteristic equation comes from $\frac{dx}{y^2(x-y)} = \frac{dy}{x^2(x-y)} $ which solution is :
$$y^3-x^3=c_1$$
This is what you rightly found : your equation (1).
A second characteristic equation comes from $\frac{dx}{y^2(x-y)} = \frac{dz}{z(x^2 + y^2)}$
$$\frac{dz}{z}=\frac{x^2+y^2}{y^2(x-y)}dx$$
with $y=(c_1+x^3)^{1/3}$
$$\frac{dz}{z}=\frac{x^2+ (c_1+x^3)^{2/3} }{(c_1+x^3)^{2/3}(x-(c_1-x^3)^{1/3})}dx$$
$$z=c_2\exp\left(\int \frac{x^2+ (c_1+x^3)^{2/3} }{(c_1+x^3)^{2/3}(x-(c_1+x^3)^{1/3})}dx \right)$$
The difficulty is to integrate for a closed form. If we could do it we would have a function $I(c_1,x)$ such as :
$$I(c_1,x)=\int \frac{x^2+ (c_1+x^3)^{2/3} }{(c_1+x^3)^{2/3}(x-(c_1+x^3)^{1/3})}dx $$
and a second characteristic equation would be :
$$z\,\exp(-I\left(c_1,x \right) )=c_2$$
$$z\,\exp(-I(y^3-x^3,x) )=c_2$$
General solution of the PDE expressed on implicit form $c_2=F(c_1)$ :
$$z=\exp(I(y^3-x^3,x) ) F\left(y^3-x^3 \right)$$
where $F$ is an arbitrary function to be determined according to some conditions (not specified in the wording of the question).
Nevertheless the solution is not obtained on a closed form, due to the weird integral.
