Answer
Let $u=(u_1,u_2)=(0,1)$, then $\langle u.u\rangle=3u_1u_2-u_1u_2=2u_1u_2$ and hence $\langle (0,1),(0,1)\rangle=0$ but $u=(0,1)\neq 0$ . This shows that the function does not define an inner product.
Work Step by Step
Let $u=(u_1,u_2)=(0,1)$, then $\langle u.u\rangle=3u_1u_2-u_1u_2=2u_1u_2$ and hence $\langle (0,1),(0,1)\rangle=0$ but $u=(0,1)\neq 0$ . This shows that the function does not define an inner product.