Metric and Convariant Tensor 
$g_{ij}$ is the metric tensor. Show that $g^{ij}$ which satsifies $g_{ij}g^{jk}=\delta_i^k$ is a covariant tensor of rank $2$.

I am not sure how to show this? Does it instead mean to show that $g_{ij}$ is a convariant tensor of rank $2$?
 A: Well, you can take the equation
$$
g_{ij}g^{jk}=\delta_i^k
$$
as a definition of $g^{jk}$ and establish its transformation properties. 
This (tensor) equation should be valid in any basis, so
$$
g'_{ij}{g'}^{jk}={\delta'}_i^k = {\delta}_i^k
$$
The change of basis formula for the covariant components is 
$$
g'_{ij} = \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} g_{pq}
$$
So in your new basis,
$$
\frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} g_{pq} {g'}^{jk} = {\delta}_i^k = g_{ij} {g}^{jk}
$$
or
$$
g_{pq}\left( \delta^p_{i}\delta^q_{j} g^{jk} - \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} {g'}^{jk} \right) = 0 \\
\delta^p_{i}\delta^q_{j} g^{jk} = \frac{\partial x_p}{\partial x'_i}\frac{\partial x_q}{\partial x'_j} {g'}^{jk}
$$
Contracting both sides with $\delta^i_k$ and simplifying, you'll have
$$
g^{qp} = \frac{\partial x_p}{\partial x'_k}\frac{\partial x_q}{\partial x'_j} {g'}^{jk}
$$
You can always invert this by contracting both sides with $\dfrac{\partial x'_n}{\partial x_p}\dfrac{\partial x'_m}{\partial x_q} $ to get a more familiar form in which the coordinates after a change of basis are on the left hand side.
$$
{g'}^{mn} = \frac{\partial x'_m}{\partial x_q}\frac{\partial x'_n}{\partial x_p} g^{qp}
$$
which is the transformation law for the contravariant components of the 2nd order tensor.
