To Solve: $$\displaystyle x(z-2y^2)\frac{\partial z}{\partial x}=\left(z-\frac{\partial z}{\partial y}\right)(z-y^2-2x^3)$$
Forming the subsidiary equations: $\displaystyle \frac{dx}{x(z-2y^2)}=\frac{dy}{yz-y^3-2x^3y}=\frac{dz}{z^2-y^2z-2x^3z}$
My Attempt: $\displaystyle \frac{dy}{y(z-y^2-2x^3)}=\frac{dz}{z(z-y^2-2x^3)}$
$\displaystyle \frac{dy}{y}=\frac{dz}{z}$
$\displaystyle \frac{y}{z}=C_1 ; d\left(\frac{y}{z}\right)=0$
I can't think of the second part .. Please help.
The given answer is: $\displaystyle f\left(\frac{y}{x},\frac{z}{x}-\frac{y}{x}+x^2\right)=0$
Please take a look at the solution of this question and this question
The goal is to prove that $$d\left(\frac{y}{x}\right)=0,d\left(\frac{z}{x}-\frac{y}{x}+x^2\right)=0$$