Find all values of $n$ such that $\varphi(n) = n/6$. Using the product formula (the formula with the prime factors of $n$), I got
$$1=6\frac{(P_1-1)}{P_1}\frac{(P_2-1)}{P_2}\cdots\frac{(P_k-1)}{P_k}\,.$$
i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $\frac{p-1}{p}.$ 

There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $\frac{1}{6}$ byt choosing many large primes.

However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p&lt;q,$ we also have $p-1 &lt; q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$ 

For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ \{ \}, \frac{1}{1}   $$
$$ \{ 2\}, \frac{1}{2}   $$
$$ \{ 3\}, \frac{2}{3}   $$
$$ \{ 2,3\}, \frac{1}{3}   $$
Hints:

We know that if $\;p_1,...,p_k\;$ are the different primes diving $\;n\;$ , then

$$\varphi(n)=n\prod_{j=1}^k\left(1-\frac1{p_j}\right)$$

Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.