is $F(x, n)=x(n)u(n)$ time invariant? My book with answers to proakis and manolakis book says that $F(x,n)=x(n)u(n)$ is time invariant while my calculation shows that it is time variant. I have repeated them many times and I am mostly convinced that book is wrong. Can somebody please check my calculations?  

let $u(n)$ be a Heavside step function. First we shift output signal:

$$F(x, n-k)=x(n-k)u(n-k)$$

Let $x'(n)=x(n-k)$ be a shifted input signal. Lets calculate response to shifted input signal:

$$F(x', n)=x'(n)u(n)=x(n-k)u(n)$$

since $F(x, n - k) \neq F(x', n)$ this system is time variant. Is my method correct?
The answer to this (and your other similar question) is most likely found in the somewhat unusual notation used for the system.

In ordinary terms, if you just take your system as a map between input and output signal, then you're absolutely right, it would not be time invariant.

However, Proakis defines the system as a map from an input signal and a time coordinate to the output signal. If you want, the time coordinate itself can be understood as a second signal that just happens to contain the time index at each time index. Consequently, the time shift operator would act on this system definition by both shifting the proper signal and the time coordinate, so that the the system turns out to be time invariant.

So let me summarise: The system map $F(x,n)=x(n)u(n)$ is time invariant whereas the map $F(x)=x(n)u(n)$ is not time invariant.