Let $f : U \rightarrow V$ be a smooth map between open sets $U \subset \mathbb{R}^{m}$ and $V \subset \mathbb{R}^{n}$. Let us write $f=\left(f_{1}, \ldots, f_{n}\right)$ and let's use different variables for domain and codomain $\left(x^{1}, \ldots, x^{n}\right) \in \mathbb{R}^{n}$ e $\left(y^{1}, \ldots, y^{m}\right) \in \mathbb{R}^{m}$. I am asked to prove that $$f^{*}\left(d x^{i}\right)=\frac{\partial f_{i}}{\partial y^{j}} d y^{j}=d f_{i}$$ Let $v \in T_p \mathbb{R}^{n} = \mathbb{R}^{n}$. My proof goes like this $$\left(f^{*}\left(d x^{i}\right)\right)(p) (v) = d x^{i}_{f(p)} (df_p(v)) = d (x^{i}\circ f)_p(v) = d (f_i)_p (v)$$ and hence $$f^{*}\left(d x^{i}\right) = d f_{i} = \frac{\partial f_{i}}{\partial y^{j}} d y^{j}$$
However, I am supposed to prove $f^{*}\left(d x^{i}\right)=\frac{\partial f_{i}}{\partial y^{j}} d y^{j}$ directly. I managed to get $\forall p \in U \ \forall v \in T_p U \ \forall g \in C^\infty (\mathbb{R})$
\begin{align} \left(f^{*}\left(d x^{i}\right)\right)(p) (v) (g) & = d x^{i}_{f(p)} \left(df_p \left(\sum_j v^j \frac{\partial}{\partial y^j} \right) \right) (g) \\ & = \sum_j v^j \left( d (f_i)_p \left( \frac{\partial}{\partial y^j} \right) \right) (g) \\ & = \sum_j v^j \frac{\partial g \circ f_i}{\partial y^j} (p) \\ & = \sum_j dy^j_p(v) g'(f_i(p)) \frac{\partial f_i}{\partial y^j} (p) \end{align}
but I'm not really sure whether this is right. Does anyone know how to go on from here?