Differential forms and homogeneous functions The question is: Let be g: $\mathbb{R}^{3}  \rightarrow  \mathbb{R}$ homogeneous of degree k, i.e, $g(tx,ty,tz) = t^{k}g(x,y,z), t>0$ and $(x,y,z) \in \mathbb{R}^{3}$.
a) Prove that $xg_{x} + yg_{y} + zg_{z} = kg$ (The Euler's relation). This part is ok ! but:
b) Let be $ w = adx + bdy + cdz $ where a, b and c are homogeneous of degree k and $dw = 0$ . Show that $w = df$ where $$ f = \frac{xa + yb + zc}{k+1}$$
The hint is : notice that $dw=0$ implies $ b_{x} = a_{y} , c_{x} = a_{z} , b_{z} = c_{y}$ (Partial Derivaties) and apply Euler's relation.
Well, i tried to do the follow:
$$ df = \frac{1}{k+1} d(xa + yb + zc) = $$ (omitting the denominator) $d(xa) + d(yb) + d(zc),$ and $d(xa) = \sum  \frac{\partial xa}{\partial x_{i}}dx_{i} = adx + a_{xx}dx + a_{xy}dy + a_{xz}dz$ 
$d(yb) = \sum  \frac{\partial yb}{\partial x_{i}}= bdy + b_{yx}dx + b_{yy}dy + b_{yz}dz$
$d(zc) = cdz + c_{zx}dx + c_{zy}dy + c_{zz}dz$
Is this alright ? 
Now, how can i use the hint ?
Thank's !
 A: Write $\omega = \sum a^i dx_i$ where the $a^i$ are homogeneous, and let $F =\sum x_i a^i$. Then 
$$dF = \sum dx_i a^i +\sum x_ida^i = \omega +\sum x_ida^i,$$
so we need to check that the sum $\sum x_ida^i$ equals $k\omega$ under the hypothesis that $d\omega=0$.  With the suggestive notation $x\cdot da = k\omega$, this is a generalization of Euler's formula for closed forms that are homogeneous. 
Now, by a use of Euler's formula for the $a^i$, we see that $$k\omega = \sum ka^i dx_i = \sum \frac{\partial a^i}{\partial x_j} x_j dx_i,$$
so that
\begin{align*} \sum x_ida^i - k\omega &=\sum x_i  \frac{\partial a^i}{\partial x_j}dx_j -  \sum \frac{\partial a^i}{\partial x_j} x_j dx_i \\
&=\sum \left( \frac{\partial a^i}{\partial x_j}-\frac{\partial a^j}{\partial x_i}\right)x_i  dx_j 
\end{align*}
Finally, the hypothesis that  $d\omega=0$ means the following sum is zero
$$\sum \frac{\partial a^i}{\partial x_j} dx_i\wedge dx_j$$
and this is exactly the claim that, for each $i<j$, we have
$$\frac{\partial a^i}{\partial x_j}=\frac{\partial a^j}{\partial x_i}.$$
