Given the equation in partial derivatives $$xz_x+2yz_y=4yx^2z$$
I am asked to find all implicit and explicit solutions of it. Following a set of indications, I found the function $$F(x,y,z)=\frac{z}{e^{yx^2}}$$ which is a first integral of the characteristic sytem of equations: $$\begin{cases}x'=x\\y'=2y\\z'=4yx^2z\end{cases}$$meaning that $xF_x+2yF_y+4yx^2zF_z=0$.
How do I find all the implicit and explicit solutions from here?
There is a mistake in your calculus since you found $xF_x+2yF_y+4yx^2zF_z=0$ which is not $xF_x+2yF_y-4yx^2zF_z=0$ as expected. Without the details of your calculus one cannot say where exactly the mistake of sign occurred.
The system is OK : $\quad\begin{cases}\frac{dx}{ds} =x\\\frac{dy}{ds} =2y\\\frac{dz}{ds} =4yx^2z\end{cases}\quad$ or on an equivalent form : $$ds=\frac{dx}{x}=\frac{dy}{2y}=\frac{dz}{4x^2yz}$$ A first characteristic equation comes from $\frac{dx}{x}=\frac{dy}{2y}$ : $$\frac{y}{x^2}=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{dy}{2y}=\frac{(2xy)dx+(x^2)dy}{(2xy)x+(x^2)2y}=\frac{d(x^2y)}{4x^2y}=\frac{dz}{4x^2yz}$
$d(x^2y)=\frac{dz}{z}\quad;\quad x^2y=\ln|z|+$constant. $$z\:e^{-x^2y}=c_2$$ The general solution of the PDE expressed on the form of implicit equation is : $$\Phi\left(c_1(x,y,z),c_2(x,y,z)\right)=0$$ $\Phi$ is an arbitrary function of two variables. $$\Phi\left(\frac{y}{x^2}\:,\:\:ze^{-x^2y}\right)=0$$ Of course one can write $\quad\phi\left(\frac{y}{x^2}\:,\:\:ze^{-x^2y}\right)=C\quad$ with $\phi=\Phi+C$
Equivalently : $\quad c_2(x,y,z)=F\left(c_1(x,y,z) \right)$ $$ze^{-x^2y}=F\left(\frac{y}{x^2}\right)$$ $F$ is an arbitrary function, to be determined according to some boundary condition (not specified in the wording of the question). The explicit general solution is : $$z(x,y)=e^{x^2y}F\left(\frac{y}{x^2}\right)$$