I have to find the curvature (sectional, of Ricci and scalar) of $\mathbb S^n\subset \mathbb R^{n+1}$. My formula are
For the sectional curvature:
$$K_p(\pi)=\frac{R(X,Y,Y,X)}{\|X\wedge Y\|^2}$$ where $\pi\subset T_pM$ is a $2-$plan, $\{X,Y\}$ a basis of $\pi$ and $$R(X,Y,Z,W)=\left<R_{XY}Z,W\right>$$ where $R_{XY}Z=[\nabla _X,\nabla _Y]Z-\nabla _{[X,Y]}Z.$
Ricci
$$Ric(X,Y)=Trace(Z\longmapsto R_{ZX}Y)=\sum_{i=1}^nR(e_i,X,Y,e_i)$$ where $e_i$ is a orhonormal basis of $T_pM$.
Scalar curvature
$$S(p)=Tr(Ric_p)=\sum_{i=1}^nRic(e_i,e_i).$$
Even if I have the formula, I absolutely don't know how to apply them in the context.
Probably for $\mathbb S^2\subset \mathbb R^3$, I can do as follow. $$T_pM=Span(\partial _\theta,\partial _\varphi)$$ where $$\partial _\theta=(-\sin\theta\sin\varphi,\cos\theta\sin\varphi,0)$$ $$\partial _\varphi=(\cos\theta\cos\varphi,\sin\theta\cos\varphi,-\sin\varphi),$$ but this is indeed orthogonal, but not othonormal basis. Do I have to do as follow ? And if yes, it's long computation, is there easy way to compute this ?
Yes, for $\mathbb{S}^2$ you can do it like that (but be careful to the poles!)
Another possibility is to notice that for every two points $p,q$ there is an isometry of the sphere sending $p$ to $q$, so the curvature must be constant, say $K$, on the whole sphere. Then the Gauss-Bonnet theorem tells you that $$4\pi K = \int K dA = 4\pi.$$ Therefore, $K=1$.
Similar arguments can help you for the higher dimensional spheres, too. However, for general manifolds you usually have to do all the computations.