Answer
see the counterexample below.
Work Step by Step
Let $u=(1,1,2), v=(0,1,0), w=(0,1,1)$. Now, using the definition of the function we have
$$\langle u,w\rangle=\langle (1,1,2),(0,1,1)\rangle=3, $$
$$\langle v,w\rangle=\langle (0,1,0),(0,1,1)\rangle=0 $$
$$\langle u+v,w\rangle=\langle (1,2,2),(0,1,1)\rangle=4,$$
One can see that $$\langle u+v,w\rangle\neq \langle u,w\rangle+\langle v,w\rangle .$$
Hence, the function does not define an inner product.