I need to solve equation:
$$\dfrac{d}{dz}(p_0p_1)=-\dfrac{8}{0.7+0.3z}\dfrac{dp_0}{dz}$$
If I multiply whole equation with $dz$, I have:
$$d(p_0p_1)=-\dfrac{8}{0.7+0.3z}dp_0$$ and integration of second equation does not take $z$ into integration, so solution is: $$p_0p_1=-\dfrac{8}{0.7+0.3z}p_0+C$$
Is it correct, because I had differentials according to $z$ in first shape of my equation? If it is not, my problem is what would be correct way of integration?
On
$$\dfrac{d}{dz}(p_0p_1)=-\dfrac{8}{0.7+0.3z}\dfrac{dp_0}{dz}$$ The next line is correct but useless since you have still a z in the equation and cant get rid of it so.. If $p_1$ does not depend on z then $$p_1\dfrac{d}{dz}(p_0)=f(z)\dfrac{dp_0}{dz}$$ $$(f(z)-p_1)\dfrac{dp_0}{dz}=0$$ $$\implies p_0 \text { is constant }$$
If $p_1$ depends on z $$(f(z)-p_1)\dfrac{dp_0}{dz}=p_0\dfrac{dp_1}{dz}$$ you need more informations ...a link between $p_1$ and z or $p_0$ and z
$$\dfrac{d}{dz}(p_0p_1)=-\dfrac{8}{0.7+0.3z}\dfrac{dp_0}{dz}$$ $$d(p_0p_1)=-\dfrac{8}{0.7+0.3z}dp_0 \quad \text{is correct}.$$ You wrote : $\quad d(p_0p_1)=-\dfrac{8}{0.7+0.3z}dp_0\quad$ and integration of second equation does not take $z$ into integration. This is NOT correct.
Obviously $\dfrac{dp_0}{dz}\neq 0$ in your first equation (if not, the equation would be trivial). Thus $p_0$ is function of $z$ and in return $z$ is function of $p_0$. In other words $z$ and $p_0$ are related.
So, one cannot integrate $\quad\int\dfrac{8}{0.7+0.3z}dp_0\quad$ in considering that $z$ acts as a constant relatively to the variable $p_0$ , as you did.
As a consequence $\quad p_0p_1=-\dfrac{8}{0.7+0.3z}p_0+C\quad$ is false.
If $p_1$ is also function of $z$ the problem is undetermined because they are two unknown functions $p_0(z)$ and $p_1(z)$ but only one equation. Thus one equation is missing in the wording of the problem.