How compute $\pi_3(S^2)$ ? My question is described well in the title. 
As you know, $\pi_3(S^2) = {\bf Z}$ 

One possible approach is via the long exact sequence of homotopy groups as applied to the Hopf fibration $S^1\rightarrow S^3\rightarrow S^2$.  I'm assuming you already know $\pi_k(S^k)\cong \mathbb{Z}$ and $\pi_k(S^1) = 0$ for $k\geq 2$.
The LES looks like $$...\rightarrow \pi_k(S^1)\rightarrow \pi_k(S^3)\rightarrow \pi_k (S^2)\rightarrow \pi_{k-1}(S^1)\rightarrow ...$$ and applied to any fibration, not just the Hopf fibration.
From this, using the fact that $\pi_k(S^1) = 0$ for $k\geq 2$, we learn that $\pi_k(S^3)\cong \pi_k(S^2)$ for all $k\geq 3$.  In particular, $\pi_3(S^2)\cong\pi_3(S^3) \cong \mathbb{Z}$.
