If $f\in C^k(\mathbb R^n\setminus \{0\})$ is homogeneous of degree $d$. Prove that $\partial_x^{\alpha}f$ is homogeneous of degree $d-|\alpha|$ for all $|\alpha|\leq k$.
By the definition of homogeneous, i need to show that $$\partial_x^{\alpha}f(rx)=r^{d-|\alpha|}\partial_x^{\alpha}f(x)$$.
So, my attempt was to do for the simple's case
$\partial_j f(rx)=r^d\partial_j f(x)$ because $f$ is homogeneous of degree $d$. The other side we have too that $\partial_j f(rx)= (\partial_j) f(rx)\partial_j (rx)$. I would like that $\partial_j (rx)=r$ because in this case if i organize then $\partial_j f(rx)=r^{d-1}\partial_j f(x)$ but for me not have sense $\partial_j (rx)=r$ because $rx=(rx_1,rx_2,\ldots,rx_n)$ (Am i ok in this step?).
My second idea after fixing the first step is let suppose that the statement is true for $|\alpha|=k$ and i will prove for $|\tilde{\alpha}|=k+1$ in this case i guess that $\tilde{\alpha}=\alpha +1$ and here i do not how to continued. Please i would prove my two steps please, any help please?
Thanks
Let $f_r(x) = f(rx)$. On one hand, for each $j=1,\dots,n$, $$\frac{\partial^{\alpha_j }f_r}{\partial x_j^{\alpha_j}}(x)= r^{\alpha_j}\frac{\partial^{\alpha_j }f}{\partial x_j^{\alpha_j}}(rx)$$ by the chain rule. Hence, $$D^\alpha f_r(x) = r^{\vert \alpha \vert } D^\alpha f(rx). \tag{$\ast$} $$ On the other hand, by homogeneity, $f_r(x) = r^d f(x)$, so $$ D^\alpha f_r(x)=r^d D^\alpha f(x). \tag{$\ast\ast$}$$ From ($\ast$) and ($\ast\ast$), it follows that $$D^\alpha f(rx) = r^{d-\vert \alpha \vert}D^\alpha f(x) $$ as required.
Edit: I decided to answer your question in the comments here since there is more room.
First of all you have some serious notational issues: the notation $\partial_j f (rx)$ mean that you calculate $\partial_jf$ then evaluate it at the point $rx$. On the other hand, $(\partial_j) f(rx)$ isn't really a notation I've seen but read (to me at least) as $\partial_j[f(rx)]$, which mean that you should in fact have $(\partial_j)f(rx) = r \partial_j f(rx)$ not the other way around. This is why I used the notation $f_r(x)$ above because it makes everything unambiguous.
Secondly, it doesn't appear that you are using the chain rule correctly. The line "$\partial_j f(rx) = (\partial_j)f(rx) \partial_j(rx)$" (barring the notation issues mentioned above) doesn't make sense since the left hand side is a scalar and the right hand side is a vector. Recall, if $\phi : \mathbb R^n \to \mathbb R^n$ then the chain rule says that $$\partial_j [ (f \circ \phi)(x)] = \sum_{k=1}^n \partial_k f(\phi(x)) \partial_j\phi_k(x). $$ In your case, $\phi(x) = rx$, so $\partial_j\phi_k(x)=r$ if $k=j$ and $\partial_j\phi_k(x)=0$ if $j\neq k$. Hence, the chain rule gives $$\partial_j [ f (rx)] = r \partial_j f(rx) . $$ To obtain $\frac{\partial^{\alpha_j }f_r}{\partial x_j^{\alpha_j}}(x)$, you just repeat this $\alpha_j$ times.