I'm learning the method of characteristics. Suppose we want to find
$$u(t,x)$$
such that
$$u = xu_x + u_t$$
$$u(0,x) = f(x)$$
By the multivariable chain rule, and putting the PDE again below to compare:
$$\frac{du}{dt} = \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial t}\frac{dt}{dt}$$
$$u = xu_x + u_t$$
By comparsion we get:
$$\frac{du}{dt} = u \rightarrow u = K_1e^t$$
$$\frac{dx}{dt} = x \rightarrow x = K_2e^t$$
So, at the path $(t,K_2e^t)$ we have that $u$ is $K_1e^t$. But there is a family of paths $(t,K_2e^t)$ because $K_2$ is undetermined. Let's see the graph for one possible $K_2$:
If I knew $K_2$ then I'd have this path in the graph, and then I'd know that on this path, $u(0,K_2) = f(K_2) = K_1$ so the solution for $u$ on the path $(t,K_2e^t)$ should be $u = K_1e^t = f(K_2)e^t$. However this solution is not for every $x$ and every $t$, it's just in the path, that is:
$$u(t, K_2e^t) = f(K_2)e^t$$
there's still work needed to generalize it for $u(t,x)$ in general.
So the two questions are: how to find $K_2$ first, and how to transform $u(t, K_2e^t) = f(K_2)e^t$ into a solution dependent of $t$ and $x$, that is, $u(x,t)$?
$$xu_x+u_t=u$$ Your approach is equivalent to considering the system of characteristic ODEs : $$\frac{dx}{x}=\frac{dt}{1}=\frac{du}{u}$$ A first family of characteristic curves comes from $\frac{dt}{1}=\frac{du}{u}$ leading to : $$ue^{-t}=K_1$$ A second family of characteristic curves comes from $\frac{dx}{x}=\frac{dt}{1}$ leading to : $$xe^{-t}=K_2$$ Thus the general solution of the PDE expressed on the form of implicit equation is : $$\Phi(K_1,K_2)=\Phi(ue^{-t},xe^{-t})=0$$ where $\Phi$ is an arbitrary function of two variables.
Equivalently, expressed on explicit form : $$ue^{-t}=F(xe^{-t})$$ where $F$ is an arbitrary function. $$u=e^tF(xe^{-t})$$ The function $F$ has to be determined to fit the boundary condition.
CONDITION : $$u(0,x)=f(x)=e^0F(xe^{-0})=F(x)$$ Thus the function $F$ is the same function as the known function $f$. Puting it into the above general solution leads to the particular solution which fits the boundary condition : $$u(t,x)=e^tf(xe^{-t})$$