I got the same expression but I got a value of 4-2/3 Ω. But I did it in my head and now realize that I made a mental math error (namely thinking that 8/6 is 3/2 and not 4/3). So I agree that it is 60/11 Ω.Using the TS's image, here's how I get Rth:
View attachment 111864
Removing Rl, replacing the voltage sources with shorts and starting from the left: I have 2Ω in parallel with 4Ω; then 4Ω in series; then that's in parallel with 2Ω and finally that's in series with 4Ω. It can be shown as (2Ω||4Ω+4Ω)||2Ω+4Ω; I get 60/11 Ω for this.