I'm trying to solve the PDE $$\frac{\partial \rho}{\partial t}-u(t,x)\frac{\partial \rho}{\partial x}=\rho\frac{\partial }{\partial x}u(t,x)$$ for $\rho:[0,1]\times\mathbb{R}\to \mathbb{R}$ for a given $u:[0,1]\times\mathbb{R}\to \mathbb{R}$.
By the method of characteristics we get $$dt=\frac{dx}{-u}=\frac{d\rho}{\rho u_x}$$
So $$dx=-udt,$$ $$\frac{d\rho}{\rho}=\frac{u_x}{-u}dx\Rightarrow \rho=e^{A\int \frac{u_x}{-u}dx}$$ and $$\frac{d\rho}{\rho}=u_x dt\Rightarrow \rho=e^{B\int u_xdt}.$$ I am not sure if this is right, and if it is, I have no idea how to proceed.
As you wrote, the Lagrange-Charpit equations for the PDE $$ \rho_t-u(t,x)\rho_x=\rho u_x(t,x) \tag{1} $$ are given by $$ dt=\frac{dx}{-u}=\frac{d\rho}{\rho u_x}. \tag{2} $$ In the special case in which $u$ has the form $u(t,x)=w(t)v(x)$, we can rewrite $(2)$ as $$ dt=\frac{dx}{-w(t)v(x)}=\frac{d\rho}{\rho w(t)v'(x)}. \tag{3} $$ The first equality is now a separable ODE: $$ dt=\frac{dx}{-w(t)v(x)}\implies \int_{0}^t w(s)\,ds=-\int_{0}^x\frac{dz}{v(z)}+C_1. \tag{4} $$ The second equality in $(3)$ is also a separable ODE: $$ \frac{dx}{-w(t)v(x)}=\frac{d\rho}{\rho w(t)v'(x)} \implies \int\frac{d\rho}{\rho}=-\int\frac{v'(x)\,dx}{v(x)} $$ $$ \implies \ln|\rho|=-\ln|v(x)|+C_2 \implies \rho=\frac{C_2}{v(x)}. \tag{5} $$ From $(4)$ and $(5)$ it follows that $$ \rho(t,x)=\frac{1}{v(x)}F\left(\int_{0}^t w(s)\,ds+\int_{0}^x\frac{dz}{v(z)}\right), \tag{6} $$ where $F$ is an arbitrary differentiable function.