<s>[INST] Solving $3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$, I keep getting the same wrong answer For the conditions  $x\left(t\right)=t^{\frac{1}{2}},\:y\left(t\right)=t^{\frac{1}{3}},\:z\left(t\right)=0,\:t>0$
I get this:
$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
$$(I)\:\:\: \frac{dx}{3y^2}=\frac{dz}{3y^2} $$
$$x=z+C_1$$
$$C_1 = x-z$$
$$(II)\:\:\: \frac{dy}{-z+x}=\frac{dz}{3y^2} $$
$$3y^2dy\:=\:\left(-z+x\right)dz$$
$$y^3=-\frac{z^2}{2}+zx+C_2$$
$$C_2=y^3+\frac{z^2}{2}-zx$$
Putting in conditions
$$C_1=t^{\frac{1}{2}}$$
$$C_2=t$$
$$C_2=C_1^2$$
This means that
$$y^3+\frac{z^2}{2}-zx = \left(x-z\right)^2$$
$$2y^3+z^2-2zx = 2 \left(x-z\right)^2$$
$$2y^3+\left(z^2-2zx+x^2-x^2\right)\:=\:2\:\left(x-z\right)^2$$
$$2y^3-x^2+\left(x-z\right)^2\:=\:2\:\left(x-z\right)^2$$
$$2y^3-x^2=\:\left(x-z\right)^2$$
$$\sqrt{2y^3-x^2}=x-z$$
$$x-\sqrt{2y^3-x^2}=z$$
But it's still wrong
So where's the mistake?
 [/INST] $$\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
You should use the constant $C_1$ instead. Since $C_1=x-z$. Your integration of this differential equation is  not correct.
$$3y^2{dy}=C_1{dz}$$
Integrate.
$$y^3=C_1z+C_2$$
$$y^3=(x-z)z+C_2$$
$$C_2=y^3-xz+z^2$$

This line seems not correct to me:
$$y^3=-\frac{z^2}{2}+zx+C_2$$
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