How can I actually solve this kind of partial differential equations? $$ x \frac{\partial z}{\partial x}+t \frac{\partial z}{\partial t}+y \frac{\partial z}{\partial y}=xyt$$
I can see that one soultion for this equation is $$z=(1/3)xyt+ C$$ however how can one solve this kind of problems within a proper way and reach more general solutions?
 A: You have found one particular solution, and since the PDE is linear any other solution will differ from that one by a solution of the homogeneous equation:
$$
x \frac{\partial z}{\partial x}+t \frac{\partial z}{\partial t}+y \frac{\partial z}{\partial y}=0
.
$$
This equation says that the directional derivative of $z$ in the radial direction is zero; in other words, $z$ is constant on each ray from the origin.
So the general solution of your PDE is
$$
z(x,y,t) = \frac{xyt}{3} + g(x,y,t)
$$
where $g$ is a function which depends only on the direction of the vector $(x,y,t)$, not on its magnitude. (Expressed in spherical coordinates, $g$ would depend only on the angles $\theta$ and $\phi$, not on the radial variable $r$.)
There are some issues to consider if you want a smooth function at the origin ($g$ must be constant in that case), but this is to be expected since the left-hand side of your PDE is zero there.
A: $$
x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+t \frac{\partial z}{\partial t}=xyt$$
Solving thanks to the method of characteristics :
The characteristic equations are :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dt}{t}=\frac{dz}{xyt}$$
From $\frac{dx}{x}=\frac{dt}{t}$ the first characteristic : $\frac{x}{t}=c_1$
From $\frac{dy}{y}=\frac{dt}{t}$ the second characteristic : $\frac{y}{t}=c_2$
From the combination of the characteristic equations, 
$\frac{dx}{x}=\frac{dy}{y}=\frac{dt}{t}= \frac{ytdx+xtdy+xydt}{ytx+xty+xyt}=\frac{ytdx+xtdy+xydt}{3xyt}=\frac{d(xyt)}{3xyt} =\frac{dz}{xyt}$
$dz=\frac{d(xyt)}{3} \quad\to\quad$  the third characteristic : $ \quad z-\frac{xyt}{3}=c_3$
Thus, the general solution on implicit form is :
$$\Phi\left(\frac{x}{t} \:,\: \frac{y}{t} \:,\: (z-\frac{xyt}{3})\right)=0$$
where $\Phi$ is any differentiable function of three variables.
Solving it for the third variable leads to the explicit form :
$$z-\frac{xyt}{3}=F\left(\frac{x}{t} \:,\: \frac{y}{t}\right)$$
where $F$ is any differentiable function of two variables.
$$z(x,y,t)=\frac{xyt}{3}+F\left(\frac{x}{t} \:,\: \frac{y}{t}\right)$$
