very elementary proof of Maxwell's theorem
I assume the $C^1$ regularity of all functions and the fact that $f$ is never zero. I think that it can be proved that if $f$ is somewhere zero, then it is everywhere zero. The idea I came up with is the following:
$$f(x_1,\ldots,x_n)=f_1(x_1)\cdots f_n(x_n)=\phi(x_1^2+\cdots+x_n^2)$$
Denote $r^2=x_1^2+\cdots+x_n^2$. Differentiate with respect to $x_i$ and get
$$f_1(x_1)\cdots f_i'(x_i)\cdots f_n(x_n)=\phi'(r^2)2x_i$$
Divide by $f$ and obtain
$$\frac{f_i'(x_i)}{f_i(x_i)}\frac{1}{2x_i}=\frac{\phi'(r^2)}{f(x_1,\ldots,x_n)}$$
Therefore the LHS is independent of $i$. Since each $f_i$ depends only of $x_i$ it follows that there exists a constant $C_i$ such that $$\frac{f_i'(x_i)}{f_i(x_i)}\frac{1}{2x_i} =C_i$$ From here it is easy to get $f_i=C_ie^{C_i x_i^2}$ from where the conclusion follows. $\ $ $\ $
Maxwell's theorem was proved for describing distribution of particles velocities. Try to use physics intuition to explain this theorem. For example we can explain why this distribution is exponential, this way:
Distribution of kinetic energy $f(E)$ depends on distribution of kinetic energy of velocity components $\phi(E_x),\phi(E_y),\phi(E_z)$, where $E_x=\frac{m v_x^2}{2}$, $E_y=\frac{m v_y^2}{2}$, $E_z=\frac{m v_z^2}{2}$ and $E=E_x+E_y+E_z$. But this velocities are independent, so $$ f(E_x+E_y+E_z)=\phi(E_x)\phi(E_y)\phi(E_z) $$
Consider all particles with fixed kinetic energy $E$ and fixed velocity $v_z$. This particles can have different velocity components $v_x,v_y$ But since $E$ is fixed we have $$ E_x+E_y=\text{const}\tag{1} $$ $$ \phi(E_x)\phi(E_y)=\text{const} $$ The last statement can be rewritten as $$ \ln\phi(E_x)+\ln\phi(E_y)=\text{const}\tag{2} $$ Since we are use physical approach we can carelessly take differentials of (1) and (2) $$ dE_x+dE_y=0 $$ $$ \frac{\phi'(E_x)}{\phi(E_x)}dE_x+\frac{\phi'(E_y)}{\phi(E_y)}dE_y=0 $$ Then you get $$ \frac{\phi'(E_x)}{\phi(E_x)}=\frac{\phi'(E_y)}{\phi(E_y)} $$ The left hand side is a function of $E_x$, the right hand side is a function of $E_y$. This is possible if $$ \frac{\phi'(E_x)}{\phi(E_x)}=\alpha_x=\text{const} $$ The solution of this differential equation is $$ \phi(E_x)=A_x e^{\alpha_x E_x} $$ Similarly, for other components $$ \phi(E_y)=A_y e^{\alpha_y E_y} $$ $$ \phi(E_z)=A_z e^{\alpha_z E_z} $$ Since distribution of velocities $v_x$, $v_y$ and $v_z$ must be the same, we have $A_x=A_y=A_z=A$, $\alpha_x=\alpha_y=\alpha_z=\alpha$. Since probability for a particle to have very big velocity is small then $\alpha<0$ Hence $$ f(E_x+E_y+E_z)=\phi(E_x)\phi(E_y)\phi(E_z)=A^3e^{\alpha(E_x+E_y+E_z)} $$ In terms of velocities it can be rewritten as $$ f\left(\frac{m v_x^2}{2}+\frac{m v_y^2}{2}+\frac{m v_z^2}{2}\right)=A^3e^{\alpha\frac{m (v_x^2+v_y^2+v_z^2)}{2}} $$
And now an interesting fact. Maxwell derived his distribution on entrance exam! Examiner loved to give open problems in theoretical physics, and Maxwell got a task for deriving distribution of velocities of particles in the ideal gas. At the end o exam Maxwell showed a complete solution of the problem!
I only demonstrate that if $f$ is zero anywhere, then it is zero everywhere, to complement the accepted answer. So write $f(x_1,\ldots ,x_n)=f_1(x_1)\ldots f(x_n)= \phi(x_1^2+\ldots+x_n^2)$. I will suppose $\phi$ to be continuous. Assume that there is some $r$ such that $\phi(r^2)=0$.
Set $R= \inf \{ \,r \mid \phi(r^2)=0 \, \}$. Suppose $R>0 $. It is clear that $\phi(R^2)=0$ and that $f_i(y) \neq 0$ for all $f_i$ if $y < R$. We get the contradiction $$ 0 \neq f_1 \left(\frac{R}{\sqrt 2}\right)f_2\left(\frac{R}{\sqrt 2}\right)f_3(0)\ldots f_n(0) =\phi(R^2)=0.$$
On the other hand if $R=0$, then for some $f_i$, we have $f_i(0)=0$. It follows easily now that $\phi$ is identically zero.