Calculate the relativistic velocity

Mathematica, 17 bytes

+##/(1+##/9*^16)&

An unnamed function taking two integers and returning an exact fraction.

Explanation

This uses two nice tricks with the argument sequence ##, which allows me to avoid referencing the individual arguments u and v separately. ## expands to a sequence of all arguments, which is sort of an "unwrapped list". Here is a simple example:

{x, ##, y}&[u, v]

gives

{x, u, v, y}

The same works inside arbitrary functions (since {...} is just shorthand for List[...]):

f[x, ##, y]&[u, v]

gives

f[x, u, v, y]

Now we can also hand ## to operators which will first treat them as a single operand as far as the operator is concerned. Then the operator will be expanded to its full form f[...], and only then is the sequence expanded. In this case +## is Plus[##] which is Plus[u, v], i.e. the numerator we want.

In the denominator on the other hand, ## appears as the left-hand operator of /. The reason this multiplies u and v is rather subtle. / is implemented in terms of Times:

FullForm[a/b]
(* Times[a, Power[b, -1]] *)

So when a is ##, it gets expanded afterwards and we end up with

Times[u, v, Power[9*^16, -1]]

Here, *^ is just Mathematica's operator for scientific notation.

MATL, 9 bytes

sG3e8/pQ/

Try it online!

s      % Take array [u, v] implicitly. Compute its sum: u+v
G      % Push [u, v] again
3e8    % Push 3e8
/      % Divide. Gives [u/c, v/c]
p      % Product of array. Gives u*v/c^2
Q      % Add 1
/      % Divide. Display implicitly

Jelly, 9 bytes

÷3ȷ8P‘÷@S

Try it online! Alternatively, if you prefer fractions, you can execute the same code with M.

How it works

÷3ȷ8P‘÷@S  Main link. Argument: [u, v]

÷3ȷ8       Divide u and v by 3e8.
    P      Take the product of the quotients, yielding uv ÷ 9e16.
     ‘     Increment, yielding 1 + uv ÷ 9e16.
        S  Sum; yield u + v.
      ÷@   Divide the result to the right by the result to the left.