
, and

,
we want to find the range of

Notice that Whatever value v can possibly produce, that is the Range of v, becomes the range of u.
Then whatever u produces for these values, is the range of u(v(x))
So, first find range of v: clearly the domain of v is R-{0}, as v(0) makes no sense.
We check what values c can v produce:

,
this means that any c (for now) can be produced... it is enough to let x=1/c
this also means that c cannot be equal to 0 as 1/c makes no sense if c=0.
Thus the range of v is R-{0},
Now we check the range of u(v(x)) for v(x)∈R-{0},
assume we want to produce a value c, so:

since the left side is always positive or 0, for x=0, the right hand side expression must also be positive or zero, which means c-3 must be negative or 0,
thus c-3≤0; c≤3. Here recall that x in u(x) cannot be 0, so c<3.
Answer: The range of u(v(x)) is (-infinity, 3)