If $ \ \omega=f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz \ $ and $ f, \ g , \ h \ $ are smooth functions on $ \mathbb{R}^3 \ $ , then show that $ \ d(d \omega)=0 \ $.
Answer:
$$ d \omega=df \wedge dx+dg \wedge dy+dh \wedge dz$$ or $$d \omega=(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz) \wedge dx + (\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy+\frac{\partial g}{\partial z}dz) \wedge dy + (\frac{\partial h}{\partial x}dx+\frac{\partial h}{\partial y}dy+\frac{\partial h}{\partial z}dz) \wedge dz $$ or $$d (\omega)=(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}) dy \wedge dz+ (\frac{\partial f}{\partial z}-\frac{\partial h}{\partial x}) dz \wedge dx + (\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}) dx \wedge dy $$
Next I can not proceed to show $ d(d\omega)=0 \ $ anyhow ?
Help me out
Your $d\omega$ should be $$d \omega=(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}) dy \wedge dz+ (\frac{\partial f}{\partial z}-\frac{\partial h}{\partial x}) dz \wedge dx + (\frac{\partial g}{\partial x}-{\frac{\partial f}{\partial y}}) dx \wedge dy , $$ so
\begin{align} d^2 \omega & = d \Big((g_x-f_y) dx \wedge dy + (h_y-g_z)dy\wedge dz + (h_x-f_z)dx \wedge dz \Big) \\ & = (g_{xz} - f_{yz}) dz\wedge dx \wedge dy + (h_{yx}-g_{zx}) dx\wedge dy \wedge dz + (h_{xy}-f_{zy}) dy\wedge dx \wedge dz. \end{align}
Next, arrange the terms in the second equality into just one term $ (\cdots)\, dx \wedge dy \wedge dz$ and use the fact that $f_{xy}=f_{yx}$ and so on. You'll find that the terms will cancel each other.