I know that there is a lot of computations of Gaussain Curvature of $S^{2}$, but I'm trying my on way.
Just to simplify, I'm dealing with points in the north hemisphere, that is, where $z=0.$
I'm using this parametrization:
$$\mathbb{x}(u,v)=\left(u,v,\sqrt{1-u^2-v^2}\right). $$
So, $$\Bbb{x}_{u}=\left(1,0,-u(1-u^2-v^2)^{-1/2}\right), \\ \mathbb{x}_{v}=\left(0,1,-v(1-u^2-v^2)^{-1/2}\right).$$
So, the coefficients of the first fundamental form are
$$E=\frac{1-v^{2}}{1-u^2-v^2},\;F=\frac{uv}{1-u^2-v^2},\;G=\frac{1-u^2}{1-u^2-v^2} \\ \implies EG-F^{2}=1. $$
I think that, till here, there is nothing wrong.
Now, I want to find $e,g,f$.
$$\mathbb{x}_{uu}=\left(0,0,-u^{2}(1-u^2-v^2)^{-3/2}\right),\\ \Bbb{x}_{vv}=\left(0,0,-v^{2}(1-u^2-v^2)^{-3/2}\right), \\ \mathbb{x}_{uv}=\left(0,0,-uv(1-u^2-v^2)^{-3/2}\right).$$
To compute the normal $N$, we can do:
$$N=\Bbb{x}_{u}\wedge\Bbb{x}_{v}=\det\left[ \begin{array}{ccc} 1 & 0 & -u(1-u^{2}-v^{2})^{-1/2}) \\ 0 & 1 & -v(1-u^{2}-v^{2})^{-1/2}) \\ \hat{i} & \hat{j} & \hat{k} \end{array} \right] \\ \implies N=\left(\frac{u}{1-u^2-v^2},\frac{v}{1-u^2-v^2},1\right) $$
So, we have the coefficients of the second fundamental form:
$$e=\left<\Bbb{x}_{uu},N\right>=\frac{-u^2}{(1-u^2-v^2)^{-3/2}} \\ f= \left<\Bbb{x}_{uv},N\right>=\frac{-uv}{(1-u^2-v^2)^{-3/2}} \\ g=\left<\Bbb{x}_{vv},N\right>=\frac{-v^2}{(1-u^2-v^2)^{-3/2}}$$
Now I see that there is a problem. See that
$$eg-f^2=\frac{u^2v^2-u^2v^2}{(1-u^2-v^2)^{3}}=0. $$
Since $K=\frac{eg-f^2}{EG-F^2},$ this would imply $K=0$, and this is obviously wrong.
Where is my mistake?
You calculated $EG-F^2$ incorrectly. You should get $$EG-F^2=\frac{1}{1-u^2-v^2}.$$ This will change your $N$ to $$N=\left(u,v,\sqrt{1-u^2-v^2}\right).$$ After calculating \begin{align*} x_{uu}&=\left(0,0,\frac{v^2-1}{(1-u^2-v^2)^{3/2}}\right)\\ x_{uv}&=\left(0,0,\frac{-uv}{(1-u^2-v^2)^{3/2}}\right)\\ x_{uu}&=\left(0,0,\frac{u^2-1}{(1-u^2-v^2)^{3/2}}\right), \end{align*} you'll get
\begin{align*} e&=\frac{v^2-1}{{1-u^2-v^2}}\\ f&=\frac{-uv}{{1-u^2-v^2}}\\ g&=\frac{u^2-1}{{1-u^2-v^2}}. \end{align*} Now, you can conclude that $$K=\frac{eg-f^2}{EG-F^2}=1.$$