How can I show that $(\pi_1,W_1), (\pi_2,W_2)$ is a smooth atlas for $S^3$?

Let $S^3 = \{(x^1,x^2,x^3,x^4) \in \mathbb{R}^4 | (x^1)^2+(x^2)^2+(x^3)^2+(x^4)^2 = 1\}$

Let $W_1 = S^3 - (0,0,0,1);$ and $\pi_1: W_1\to \mathbb{R}^3$ by

$\pi_1(x^1,x^2,x^3,x^4) = \left (\frac{x^1}{1-x^4},\frac{x^2}{1-x^4},\frac{x^3}{1-x^4} \right ) $

Let $W_2 = S^3 - (0,0,0,-1);$ and $\pi_2: W_2\to \mathbb{R}^3$ by

$\pi_2(x^1,x^2,x^3,x^4) = \left (\frac{x^1}{1+x^4},\frac{x^2}{1+x^4},\frac{x^3}{1+ x^4} \right ) $

How can I show that $(\pi_1,W_1), (\pi_2,W_2)$ is a smooth atlas for $S^3$?

I know that in order to show this, I need to show that $\pi_1,\pi_2$ are diffeomorphisms, and that $W_1 \cup W_2 \supseteq S^3$ and $\pi_2 \circ \pi_{1}^{-1}$ is $C^\infty$.

but I don't know how to show any of these things computationally, how could I do that? Let $S^3 = \{(x^1,x^2,x^3,x^4) \in \mathbb{R}^4 | (x^1)^2+(x^2)^2+(x^3)^2+(x^4)^2 = 1\}$

Let $W_1 = S^3 - (0,0,0,1);$ and $\pi_1: W_1\to \mathbb{R}^3$ by

$\pi_1(x^1,x^2,x^3,x^4) = \left (\frac{x^1}{1-x^4},\frac{x^2}{1-x^4},\frac{x^3}{1-x^4} \right ) $

Let $W_2 = S^3 - (0,0,0,-1);$ and $\pi_2: W_2\to \mathbb{R}^3$ by

$\pi_2(x^1,x^2,x^3,x^4) = \left (\frac{x^1}{1+x^4},\frac{x^2}{1+x^4},\frac{x^3}{1+ x^4} \right ) $

How can I show that $(\pi_1,W_1), (\pi_2,W_2)$ is a smooth atlas for $S^3$?

I know that in order to show this, I need to show that $\pi_1,\pi_2$ are diffeomorphisms, and that $W_1 \cup W_2 \supseteq S^3$ and $\pi_2 \circ \pi_{1}^{-1}$ is $C^\infty$.

but I don't know how to show any of these things computationally, how could I do that?