I'm trying to solve $$u_x - 2u_y = u$$
with initial condition $u(x,0)=6e^{-3x}$. I begin by trying to find a solution of the form $$u(x,y)=X(x)Y(y)$$
substituting this into the PDE yields the relation $$\frac{-2X}{X-X'}=\frac{Y}{Y'}=\lambda$$
where $\lambda$ is a constant to be determined. The relation contains two ordinary differential equations $$(1)\enspace X\bigg(\frac{-\lambda-2}{-\lambda}\bigg)=X'$$ $$(2)\enspace Y'-\frac{Y}{\lambda}=0$$
The solutions are $$X(x)=e^{\alpha x},\enspace Y(y)=e^{\frac{y}{\lambda}}$$
where $\alpha = \frac{\lambda+2}{\lambda}$. How would one determine the constant $\lambda$ at this point? I tried substituting this solution into the initial condition, but it does not seem to be the way. Also, I have introduced $\lambda$ without first determining whether or not its explicitly positive or negative in this case, so the solution might be false.
You're almost there. Putting your solutions together, and including an arbitrary normalization constant, you have obtained the result that $$ u(x,y) = A \exp \left( - \frac{(\lambda + 2)x - y}{\lambda} \right). $$ Now plug in $y = 0$ to this form and see if you can figure out the values of $A$ and $\lambda$ needed to yield $u(x,0) = 6 e^{-3x}$.