Given a Riemannian metric $g$, we can always compute curvature of Levi-Civita connection with respect to $g$, that is $$ R_{ijkl}=\frac12(\partial_i\partial_kg_{jl}+\partial_j\partial_lg_{ik}-\partial_{i}\partial_lg_{jk}-\partial_j\partial_kg_{il})+g_{rs}(\Gamma_{ik}^r\Gamma_{jl}^s-\Gamma_{jk}^r\Gamma_{il}^s) $$
However, I find it's quite difficult to compute in fact, for example, on Poincar'e disk $\Bbb{B}^n=\{x\in\Bbb{R}^n\mid |x|<1\}$, there is a Riemannian metric $$ g=\frac{4\delta_{ij}\mathrm{d}x^i\otimes\mathrm{d}x^j}{(1-|x|^2)^2} $$
In this case I find it's very very very difficult to compute $R_{ijkl}$, is there any easy way to compute this? Thanks in advance for any help in advance.
By definition of curvature form we have $$ R_{ijkl}= E_i(\Gamma_{jk}^m g_{ml} ) -E_j(\Gamma_{ik}^m g_{ml} ) -\Gamma_{jk}^m\Gamma_{il}^n g_{mn } + \Gamma_{ik}^m \Gamma_{jl}^n g_{mn} $$
where $E_i$ is a coordinate field, which is a canonical basis on $\mathbb{R}^n$.
Further by definition of $\Gamma_{ij}^k$ we have $$ \Gamma_{ij}^k = \frac{1}{2}g^{ks} (E_ig_{js} +E_j g_{is} - E_s g_{ij} ) $$
Recall $g=fg_0$ where $f(x)= \frac{4}{(1-r^2)^2},\ r=\sqrt{\sum_{i=1}^n x_i^2}$ and $g_0$ is Euclidean metric.
Hence $f_k = E_k(f)=\sqrt{f}^3 (2x_k) $ so that we have $$ \Gamma_{ij}^k = \sqrt{f} (x_i\delta_{jk} + x_j\delta_{ik} -x_k\delta_{ij} ) $$
Note that $\sharp \{ i,j,k\}=3$ implies $\Gamma_{ij}^k=0$
For the calculation of $R_{1221}$ we have \begin{align*}R_{1221}&= E_1(\Gamma_{22}^1f)- E_2( \Gamma_{12}^1 f) -\Gamma_{22}^m\Gamma_{11}^m f +( \Gamma_{12}^m)^2 f \\&\\ E_1(\Gamma_{22}^1f)- E_2( \Gamma_{12}^1 f) &=-2\sqrt{f}^3-3f^2(x_1^2+x_2^2) \\ -\Gamma_{22}^m\Gamma_{11}^m f +( \Gamma_{12}^m)^2 f&= f^2\{ 2x_1^2+2x_2^2-\sum_{i>2}x_i2 \} \end{align*}
Hence $R_{1221}=-f^2$. Here $\{e_i=\frac{1}{\sqrt{f}} E_i\}$ is orthonormal wrt a metric $g$ so that the sectional curvature for $e_1,\ e_2$ is $-1$. Hence the metric $g$ has a const sectional curvature.