The Gaussian curvature of the sphere $ S^2=\{(x,y,z) \in \mathbb{R}^3 | x^2+y^2+z^2=1 \}$ is obtained via taking the determinant of the derivative of the map $(x,y,z) \mapsto (x,y,z), \ $ which is $1$.
I'd like to derive the same result using the formula for sectional curvature to understand if I got the hang of this:
$$k(R\cdot S^2) = \frac{\langle R(X,Y)Y, X \rangle}{\langle X,X\rangle\langle Y,Y\rangle-\langle X,Y\rangle^2} =\frac{\langle ([\nabla_X, \nabla_Y]Y-\nabla_{[X,Y]}Y, X \rangle}{\langle X,X\rangle\langle Y,Y\rangle-\langle X,Y\rangle^2} $$
If we take $X=\frac{\partial }{\partial y}, \ Y=\frac{\partial }{\partial x}, \ $ where:
$$\frac{\partial }{\partial x}=\left(1,0,\frac{-x}{\sqrt{1-x^2-y^2}} \right), \quad \frac{\partial }{\partial y}=\left(0,1,\frac{-y}{\sqrt{1-x^2-y^2}} \right)$$
If we set $n=\left(x,y,\sqrt{1-x^2-y^2} \right), \ $ then: $$\nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial x}=\frac{\partial}{\partial x}\left(1,0,\frac{-x}{\sqrt{1-x^2-y^2}}\right)-\left\langle \frac{\partial}{\partial x}\left(0,1,\frac{-x}{\sqrt{1-x^2-y^2}}\right), n\right\rangle n= $$ $$ \left(0,0,\frac{y^2-1}{(1-x^2-y^2)^{\frac{3}{2}}} \right)-n\left\langle n, \left(0,0,\frac{y^2-1}{(1-x^2-y^2)^{\frac{3}{2}}} \right)\right \rangle $$
The calculation get a bit cumbersome, but I wanted to know if I got the gist right.