Show that the section $g$ of $T^*\mathbb{R}^3 \otimes T^*\mathbb{R}^3$ defined by $g(x_1,x_2,x_3)=x_1^2dx_1^2+dx_2^2+dx_3^2$ defines a Riemannian metric on $\mathbb{R}^3 - \{x_1=0\}$ and compute the norm of $(1,1,1)$ as a tangent vector of $(1,0,0)$ relatively to this metric.
Honestly I don't know where to begin. A Riemannian metric on a manifold $M$ is a section $g$ of $T^* \otimes T^*$ which at each point is symmetric and positive definite.
Some help would be appreciated. Thanks.
The matrix representation of $g$ at the point $(x_1,x_2,x_3)$ is$$\left(\begin{array}{ccc}x_1^2&0&0\\0&1&0\\0&0&1\end{array}\right).$$It is clearly symmetric at any point, and positive definite for $x_1\neq0$.
At the point $(1,0,0)$ the metric $g$ coincides with the standard metric, and so the norm of any tangent vector with respect to $g$ is equal to its standard norm.