Attempt #2: I think the issue here is that we need to transform from the $(1, 3)$ curvature tensor to the $(0, 4)$ one. The formula for the components of the $(1, 3)$ version are given by:
$$R^l_{ijk} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \sum_r(\Gamma^r_{jk}\Gamma^l_{ir} - \Gamma^r_{ik}\Gamma^l_{jr})$$
This means that the $(1, 3)$ curvature can be written as
$$R_{ijk} = \sum_l \Gamma^l_{ijk} \partial_i$$
We must lower an index to produce the $(0, 4)$ tensor:
$$R_{ijkm} = g(\sum_l \Gamma^l_{ijk} \partial_i, \partial_m) = \sum_l \Gamma^l_{ijk} g(\partial_i, \partial_m)$$
In your example, the vectors $\partial_i$ are orthogonal and in particular you calculated:
$$R_{1221} = \Gamma^1_{122} g(\partial_1, \partial_1) = -4 \cdot 4 = -16$$
Now from your formula:
$$K=\frac{Rm(\partial_1, \partial_2, \partial_2, \partial_1)}{|\partial_1|^2 |\partial_2|^2}=\frac{Rm(\partial_1, \partial_2, \partial_2, \partial_1)}{16}$$
we see that curvature is $-1$.
$\newcommand{\Two}{\mathsf{I}\mathsf{I}}$
$\newcommand{\rD}{\mathrm{D}}$
You can use the second fundamental form and the Gauss-Codazzi equation. For embedded manifolds in a vector space, the Gauss-Codazzi equation reads:
$$R_{vw}w.v = \Two(v, v). \Two(w, w) - \Two(v,w). \Two(v,w)
$$
and the sectional curvature is
$$sec(v, w) = \frac{R_{vw}w.v}{(v.v)(w.w) - (v.w)^2}.
$$
The formula for the second fundamental form with the embedded metric is simply $$\Two(v, w) = (\rD_v\Pi)w = (\rD_w\Pi)v$$
where $\Pi$ is the metric compatible projection to the tangent space. The second fundamental form is a covariant derivative of the projection, but it just works out to that directional derivative if you use the embedded metric in a vector space.
Let $z = (x, y)$ satisfying the defining equation and $C(z) = yx^Tx-1$, then its Jacobian is
$$J:\omega=(\omega_x, \omega_y) \mapsto 2y(x^T\omega_x)+ \omega_yx^Tx
$$
or in matrix form $J$ is the $1\times (n+1)$ matrix $(2yx^T, x^Tx)$
$$\Pi(z)\omega = \omega - J^T(JJ^T)^{-1}J\omega.$$
Here, $JJ^T = |J|^2 = 4y^2x^Tx + (x^Tx)^2$
$$\Pi(z)\omega = \begin{bmatrix}\omega_x\\ \omega_y\end{bmatrix}
- \frac{2y(x^T\omega_x)+ \omega_yx^Tx}{4y^2x^Tx + (x^Tx)^2}\begin{bmatrix}2yx\\ x^Tx\end{bmatrix} $$
if $v=(v_x, v_y), w = (w_x, w_y)$ are two tangent vectors then
$$\Two(v, w) =- \frac{2v_y(x^Tw_x)+
2y(v_x^Tw_x) + 2 w_yx^Tv_x}{4y^2x^Tx + (x^Tx)^2}\begin{bmatrix}2yx\\ x^Tx\end{bmatrix}
$$
noting that since $w$ is a tangent vector, when you take directional derivative, any term that leaves the Jacobian intact will evaluate to zero. From here, you can evaluate the sectional curvature. You can simplify using $v_y =-\frac{2yx^Tv_x}{x^Tx}$.
See https://arxiv.org/abs/2307.10017 for some other examples of this type of calculation.
As this type of question appears quite often, let's give a general answer. If a hypersurface $M$ is given by a level set equation $\phi(x) = C$ in an inner product space $E$ with $\phi$ is a scalar function, let $g_{\phi}$ be its gradient and $hess_{\phi}$ be its hessian, consider as a bilinear form, so $hess_{\phi}(\xi, \eta)$ is a number for two vectors $\xi, \eta$. Then under the embedded metric
- The Levi-Civita connection is
$$\nabla_XY = D_XY + \frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}$$
Here, two vector fields $X, Y$ on $M$ are considered as vectors in $E$, and $D_XY$ is the flat connection on $E$, ie simple Lie derivative.
- The second fundamental form is $-\frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}$.
- The sectional curvature numerator is
$$R_{XY}Y.X = \frac{1}{|g_{\phi}|^2} \{hess_{\phi}(X, X)hess_{\phi}(Y,Y)- hess_{\phi}(X, Y)^2\}.
$$
Proof: For 1, the right hand side is a vector field, as differentiating $g_{\phi}^ . Y = 0$ using $X$, we get
$$hess_{\phi}(X, Y) + g_{\phi}. D_XY = 0$$
which implies $g_{\phi}. \{D_XY + \frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}\} =0$, it satisfies the usual conditions of a covariant derivative, and metric compatibility follows from
$$2Y . (D_XY + \frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi}) = 2Y . D_XY= D_X (Y. Y)
$$
as $Y.g_{\phi} = 0$.
For 2, it is well-known $\nabla_XY = \Pi D_XY$ if $\Pi$ is the metric-compatible projection, hence $\Two(X, Y) = D_XY - \Pi D_XY =
D_XY - \nabla_XY=-\frac{hess_{\phi}(X, Y)}{|g_{\phi}|^2}g_{\phi} $. The Gauss-Codazzi equation gives 3.
Best Answer
You already have an expression for the metric, you can compute its partial derivatives and evaluate them at $(0,0)$ in order to compute Christoffel symbols and derive the curvature. Hint: you should find $1$.
Your error is saying "since at $(0,0)$, $g_{ij} = \delta_{ij}$, then all partial derivatives vanish". This is basically an error that is equivalent to saying "all functions $f$ have zero derivative since at $x$, they are constant equal to $f(x)$."