In The Poor Man's Introduction to Tensors, exercise X.3:
In Cartesian coordinates, the simplest nonvanishing vector field you can write down is the constant vector field: $$ v^1 = 1 \\ v^2 = 0 \\ v^3 = 0 \\ $$
Lower the indices to get the components of the dual vector $v_i$. Obtain the components $v_\alpha$ for the dual vector field in spherical coordinates, and use the components of the inverse metric $g^{\alpha \beta}$ (see exercise X.2) to obtain the components for the vector field $v^\alpha$ in spherical coordinates. Finally, obtain the components for the vector in the unit-vector basis: $\hat{r}$, $\hat{\theta}$, $\hat{\phi}$.
I understand how to get the components of the dual vector field, and how to obtain the components for the vector in the unit-vector basis once I know the components in the partial derivative basis. What I don't know is how to get the components in spherical coordinates using using the inverse metric $g^{\alpha \beta}$.
The lowering of components is easy, since the metric tensor for Cartesian coordinates is the identity, so we have $v_1=1$ and $v_2=v_3=0$. The components of this dual vector in spherical coordinates are computed using the transformation law of
$$\bar{v}_\alpha=\frac{\partial x^\beta}{\partial\bar{x}^\alpha}v_\beta$$
where $x^\alpha$ are the Cartesian coordinates and $\bar{x}^\alpha$ are the spherical coordinates. Since we know $v_\beta$ for each $\beta$, we have
$$\bar{v}_\alpha=\frac{\partial x^1}{\partial\bar{x}^\alpha}$$
In these coordinates,
$$v_r=\frac{\partial x}{\partial r},\;\;v_\theta=\frac{\partial x}{\partial \theta},\;\;v_\phi=\frac{\partial x}{\partial \phi}$$
Under the formula
$$x=r\cos\phi\sin\theta $$
we have
$$v_r=\cos\phi\sin\theta,\;\;v_\theta=r\cos\phi\cos\theta,\;\;v_\phi=-r\sin\phi\sin\theta$$
To raise the indices, we use the inverse metric tensor. The metric tensor for spherical coordinates is
$$g_{\alpha\beta}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \end{pmatrix}$$
For $(r,\theta,\phi)$. We can raise the index by
$$\bar{v}^\alpha=g^{\alpha\beta}\bar{v}_\beta$$
but since $g$ is diagonal, we have
$$\bar{v}^\alpha=g^{\alpha\alpha}\bar{v}_{\alpha}$$
and there is no sum being taken. Since the matrix is diagonal the inverse of the matrix is just the reciprocal of the terms, so we have
$$\bar{v}^\alpha=\frac{1}{g_{\alpha\alpha}}\bar{v}_\alpha$$
As such,
$$\bar{v}^r=\cos\phi\sin\theta,\;\;\bar{v}^\theta=\frac{\cos\phi\cos\theta}{r},\;\;\bar{v}^\phi=-\frac{\sin\phi}{r\sin\theta}$$
The shorter way of getting this by using the transformation law
$$\bar{v}^\alpha=\frac{\partial\bar{x}^\alpha}{\partial x^\beta}v^\beta$$
but it requires that we know the derivative of the spherical coordinates with respect to the Cartesian coordinates and then transform these back to spherical coordinates.