I am asked to show the result specified below using the chain rule but I am struggling to understand how we apply the chain rule to this function. I am fine with applying the chain rule when we have a function of the form $g(t)=f(x_1(t),x_2(t),...,x_n(t))$ but with regards to the question below, why when we apply the chain rule do we get a term of the form $\frac{\partial f}{\partial x_k}$, when we are letting lambda vary and not $x_k$?
Hint: Define $g(\lambda)=f(\lambda x_1,\lambda x_2,\cdots,\lambda x_n)-\lambda^Df(x_1,x_2,\cdots,x_n)$. Note that this function $g(\lambda)=0$, and that $g'(\lambda)=0$. Apply Chain Rule on both sides of $g(\lambda)=0$ and see what will happen!
EDIT: $$\frac{\partial g}{\partial \lambda}=\sum_{k=1}^{n}\frac{\partial f}{\partial \lambda x_k}\frac{\partial \lambda x_k}{\partial \lambda}-D\lambda^{D-1}f(x_1,x_2,\cdots,x_n)\\ \frac{\partial g}{\partial \lambda}=\sum_{k=1}^{n}\frac{\partial f}{\partial \lambda x_k}x_k-D\lambda^{D-1}f(x_1,x_2,\cdots,x_n)$$ To see this easily, you may do the following:
Let $f(\lambda x_1,\lambda x_2,\cdots,\lambda x_n)=f(y_1,y_2,\cdots,y_n)$. For every $i$, $y_i=h(\lambda)$. Then you kind of have a relationship between them: $\lambda\rightarrow y\rightarrow f(\cdot)$.