Is a surface locally flat, when the curvature along a line is zero

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According to Spivak Vol. 2, p. 119 the following holds for a 2-dimensional manifold $M = \left\{\left(x, y, f(x, y)\right): x, y \in \mathbb{R}\right\}$: \begin{align*} K(x, y, f(x, y)) = \frac{\frac{\partial^2 f}{\partial x^2}\, \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \,\partial y}\right)^2}{\left(1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2\right)^2} \end{align*} Now, for $f(x, y) = x\cdot y^2 + x^2$ we calculate \begin{align*} K(x, y, f(x, y)) = \frac{4x-4y^2}{\left(1+y^4+4xy^2+4y^2+4x^2y^2\right)^2} = \frac{4(x-y^2)}{\left(1+4y^2+4xy^2+y^4+4x^2y^2\right)^2} \end{align*} which is zero for $x = y^2$. Are the following conclusions correct (I set $y=t, t \in \mathbb{R}$):

  • The surface around $(t^2, t, f(t^2, t))$ for any $t$ is locally isometric to $\mathbb{R}^2$
  • The surface around $(t^2, t, f(t^2, t))$ for any $t$ is flat