We consider the map $\varphi_0:\mathbb{S}^2\rightarrow\mathbb{R}^3$ such that $\varphi_0(\theta_1,\theta_2)=R(\cos\theta_1\cos\theta_2,\cos\theta_1\sin\theta_2,\sin\theta_1)$. I know that the induced riemmannian metic is
From this, how can I obtain that the second fundamental form is
and mean curvature H=-2/R?
You need to be able to compute the normal vector $n$ first, where $n$ is the unit vector along $\frac{\partial\varphi}{\partial x^1}$ x $\frac{\partial\varphi}{\partial x^2}$. Then by definition, $h_{ij}$ would be the inner product of $\frac{\partial^2\varphi}{\partial x^i\partial x^j}$ and $n$. The mean curvature is just half the trace of the Weingarten Map matrix, which can be obtained by multiplying the matrix of the second fundamental from and the inverse of the metric matrix.