let $(D,g_0)$ be the standard Poincare disk $$D=\{(x,y)\in\mathbb{R}^2|x^2+y^2<1\}$$ $$g_0(x,y)=\dfrac{4\mathrm{d} x\otimes\mathrm{d} x+4\mathrm{d} y\otimes\mathrm{d} y}{\left(1-(x^2+y^2)\right)^2}$$compute sectional curvature.
let $X=\dfrac{\partial}{\partial x},\,Y=\dfrac{\partial}{\partial y}$，for $$\nabla_{X}X=\dfrac{8}{(1-x^2-y^2)^3}\left(xX-yY\right)$$$$\nabla_{Y}X=\nabla_{X}Y=\dfrac{8}{(1-x^2-y^2)^3}\left(yX+xY\right)$$$$\nabla_YY=\dfrac{8}{(1-x^2-y^2)^3}\left(yY-xX\right)$$$$\dfrac{\partial}{\partial x}\dfrac{8x}{(1-x^2-y^2)^3}=\dfrac{8(1-x^2-y^2)}{(1-x^2-y^2)^4}+\dfrac{8x(-3)(-2x)}{(1-x^2-y^2)^4}=\dfrac{8(1+5x^2-y^2)}{(1-x^2-y^2)^4}$$ so \begin{align*} &\langle\nabla_X\nabla_YX,Y\rangle =\left\langle\nabla_X\left(\dfrac{8}{(1-x^2-y^2)^3}(yX+xY)\right),Y\right\rangle\\ = &\,\,\dfrac{8y\langle\nabla_XX,Y\rangle}{(1-x^2-y^2)^3}+\dfrac{8(1+5x^2-y^2)}{(1-x^2-y^2)^4}\langle Y,Y\rangle+\dfrac{8x\langle\nabla_XY,Y\rangle}{(1-x^2-y^2)^3} \end{align*} and \begin{align*} &\langle\nabla_Y\nabla_XX,Y\rangle =\left\langle\nabla_Y\left(\dfrac{8}{(1-x^2-y^2)^3}(xX-yY)\right),Y\right\rangle\\ = &\,\,\dfrac{8x\langle\nabla_YX,Y\rangle}{(1-x^2-y^2)^3}-\dfrac{8(1-x^2+5y^2)}{(1-x^2-y^2)^4}\langle Y,Y\rangle-\dfrac{8y\langle\nabla_YY,Y\rangle}{(1-x^2-y^2)^3} \end{align*} so \begin{align*} &\langle R_{XY}X,Y\rangle =\langle-\nabla_X\nabla_YX+\nabla_Y\nabla_XX,Y\rangle\\ =&\,\,-\dfrac{8(2+4x^2+4y^2)}{(1-x^2-y^2)^4}\langle Y,Y\rangle=-\,\dfrac{4^3(1+2x^2+2y^2)}{(1-x^2-y^2)^6} \end{align*} so the sectional curvature $$K=\dfrac{\langle R_{XY}X,Y\rangle}{|X\wedge Y|^2}=\dfrac{\langle R_{XY}X,Y\rangle}{\langle X,X\rangle\langle Y,Y\rangle}=-\,\dfrac{4(1+2x^2+2y^2)}{(1-x^2-y^2)^2}\neq -1\qquad ?????$$ I'm going to cry...
now I know where my stupid mistake is. The first formula, the connection, is $$\nabla_{Y}X=\nabla_{X}Y=\dfrac{8}{(1-x^2-y^2)^3}\left(y\dfrac{X}{\Vert X\Vert^2}+x\dfrac{Y}{\Vert Y\Vert^2}\right)$$ The other two formulas about connection are the same. Then I can compute the correct result $$K\equiv -1$$ Very grateful to Deane and Andrew D.Hwang for their guidance and encouragement !!!