A function $g$:$\mathbb{R}^3→\mathbb{R}$ is said to be homogeneous of degree $k$ if $g(tx,ty,tz)=t^kg(x,y,z)$, $t>0$. If $g$ is differentiable and homogeneous of degree $k$, then $$xg_x+yg_y+zg_z=kg$$
Prove that if the functions $a,b$ and $c$ are homogeneous of degree $k$ and the differential form $$w=adx+bdy+cdz$$ is closed, then $w=df$, where $$f=\frac{xa+yb+zc}{k+1}$$
My first approach is differentiate $f$ by $x,y,z$ respectively. $f_x=\frac{a}{k+1},f_y=\frac{b}{k+1},f_z=\frac{c}{k+1}$. I'm not sure about this but I guess, I can write $w=\frac{a}{k+1}dx+\frac{b}{k+1}dy+\frac{c}{k+1}dz$ since $w=df$.
Then replacing $g$ to $f$; $xf_x+yf_y+zf_z=x\frac{a}{k+1}+y\frac{b}{k+1}+z\frac{c}{k+1}=kf$. //(Edit1: It only equals to $f$ I guess.)
Does my attempt okay to prove $a,b$ and $c$ are homogeneous? Thanks!
(Edit2: I don't want to use $dw$ because it equals to $0$. Don't figure out how to implement it.)
Your formulas for partial derivatives of $f$ are wrong. Keep in mind that $a$, $b$ and $c$ are not constant expressions. When differentiating you should take them into account: $$ \begin{array}{c} f_x = \frac{1}{k+1}\left( (a + xa_x) + yb_x + zc_x \right) \\ f_y = \frac{1}{k+1}\left( xa_y + (b + yb_y) + zc_y \right) \\ f_z = \frac{1}{k+1}\left( xa_z + yb_z + (c + zc_z) \right) \\ \end{array} $$ Then, compute $d\omega$: $$ d\omega = (b_x - a_y)\; dx \wedge dy + (c_y - b_z)\; dy \wedge dz + (a_z - c_x)\; dz \wedge dx $$ On the other hand $d\omega=0$, so we conclude that $$ \begin{array}{c} b_x = a_y \\ c_y = b_z \\ a_z = c_x \\ \end{array} $$ For $f_x$ we have: $$ f_x = \frac{a + xa_x + yb_x + zc_x}{k+1} = \frac{a + xa_x + ya_y + za_z}{k+1}. $$ Then use the fact that $a$ is homogeneous of degree $k$.
Apply similar reasoning to $f_y$ and $f_z$ to prove the demanded identity.