Solve linear PDE: $(x^2+y^2+yz)p+(x^2+y^2-yz)q=z(x+y)$ by Lagrange's method

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I am working on the following PDE:

$(x^2+y^2+yz)p+(x^2+y^2-yz)q=z(x+y)$ where $p=\dfrac{\partial z}{\partial x}$ and $ q=\dfrac{\partial z}{\partial y}$

I set up the Lagrange's equations as below:

$\dfrac{dx}{x^2+y^2+yz}=\dfrac{dy}{x^2+y^2-yz}=\dfrac{dz}{z(x+y)}$

I tried multipliers $1,1,0$ which gave me: $\dfrac{dx+dy}{2(x^2+y^2)}=\dfrac{dz}{z(x+y)}$

On rearranging I got:

$(x+y)\dfrac{dx+dy}{x^2+y^2}=2\dfrac{dz}{z}$ But I do not know how to take it forward from here. On trying multipliers $1,-1,0$, I got :

$\dfrac{dx-dy}{2yz}=\dfrac{dz}{z(x+y)}$

On rearranging, I got

$\dfrac{(x+y)(dx-dy)}{2y}=dz$

Again I do not know how to take it forward from here. Please help me. Any help will be greatly appreciated.