from Differential Forms and Applications by Manfredo P. DoCarmo,
$d(f^* \omega)=f^*(d \omega)$,where $\omega$ is a $k$-form in $\mathbb R^m$ and $f:\mathbb R^n \to \mathbb R^m$ is a differentiable map.
let us first prove the result for $0$-form.Let $g:\mathbb R^m \to \mathbb R$ be a differentiable function that associates to each $(y_1,...y_m)\in \mathbb R^m$ that value $g(y_1,...y_m)\in \mathbb R$.then
\begin{eqnarray*} f^*(dg) &=&f^*(\sum_{i}\frac{\partial g}{\partial y_i}dy_i)\\&=& \sum_{i j}\frac{\partial g}{\partial y_i}\frac{\partial f_i}{\partial x_j}dx_j\\ &\overset{?}{=}&\sum_{j}\frac{\partial (g \circ f)} {\partial x_j}dx_j\\ &=&d(g \circ f)\\ &=&d(f^*g) \end{eqnarray*}
I have put a 'question Mark' on the step which I am not getting.
I mean how $\sum_{i j}\frac{\partial g}{\partial y_i}\frac{\partial f_i}{\partial x_j}dx_j=\sum_{j}\frac{\partial (gof)} {\partial x_j}dx_j$ please let me know..Thanks in advance