C++: Is std::move always invoked implicitly?

std::move is simply a no-op conversion from lvalue to xvalue. The only time it is "invoked implicitly", that is, lvalue is treated as xvalue, is in the return statement of a function.

Yes, there are many cases when you would want to write out "std::move". Common ones are writing move constructors, and doing things to std::unique_ptr (e.g. moving one into a vector of unique_ptrs)

rvalue references aren't directly relevant to the idea of moving from lvalues (which is what std::move does). On the contrary, they exist to bind to rvalues, which is a whole different story.