I wanna show that $Z^a$ is indeed a contravariant vector in the same way I showed that $Z_i$ is indeed a covariant vector (see attached image).This is how I define $Z^a$ : $Z^a = \frac{\partial y^a}{\partial x^i} e_i$
How would one do this in a similar fashion?
Let $\{\mathbf{b}_i\}_i$ be the vectors of a covariant basis, and being $\{\mathbf{b}^j\}_j$ the vectors of its contravariant basis, so that $\mathbf{b}_i \cdot \mathbf{b}^j = \delta_i^j$.
Let $\{ \mathbf{b}'_a\}_a$ the new basis and $T_a^i$ the coefficients of the transformation, so that $\mathbf{b}'_a = T_a^i \mathbf{b}_i$. We need to prove that the vectors of the basis $\{\mathbf{b}'^b \}_b$ must transform as $\mathbf{b}'^b = \hat{T}^b_j \mathbf{b}^j$, being $\hat{T}^a_b$ the coefficients of the inverse transformation. Let's assume that this hold and test if the relation $\mathbf{b}'_a \cdot \mathbf{b}'^b = \delta_a^b$ follows
$\mathbf{b}'_a \cdot \mathbf{b}'^b = T_a^i \mathbf{b}_i \cdot \hat{T}_j^b \mathbf{b}^j = T_a^i \hat{T}^b_j \mathbf{b}_i \cdot \mathbf{b}^j = T_a^i \hat{T}^b_j \delta_{i}^{j} = T_a^i \hat{T}^b_i = \delta_a^b$,
where the last equality holds since $\hat{T}$ is the inverse transformation of $T$.
Exploiting the linearity of the problem, it's possible to use this answer as a proof that a reciprocal basis is contravariant w.r.t the reference basis, namely that
$\mathbf{b}'_a = T_a^i \mathbf{b}_i \implies \mathbf{b}'^b = \hat{T}^b_j \mathbf{b}^j$
