Prove the Mobius function relation

Question

I have a proof of the following relation:

$$\prod_{d|n} \frac{17-(6\mu(d)+1)^2}{16}=1$$

where the $\mu(d)$ is Mobius function.

The way I have proved it is very long and even putting it here is not realistic. And I think the way I went is not the right way in terms of efficiency. Could you please help with finding a simple way to prove this relation?

EDIT: I need to add that it is true for $n>1$. Here is maple code for testing.

with(NumberTheory):

seq(simplify(1-mul(1/16*(17-(6*Moebius(d)+1)^2), d = numtheory:-divisors(n))), n = 1 .. 100)

Answer 1

$$f(n) = \frac{17-(6\mu(n)+1)^2}{16} = (-2)^{\mu(n)}$$

Thus $$\prod_{d | n} f(d) =(- 2)^{\sum_{d | n} \mu(d)} = 1 \text{ for } n > 1, = -2 \text{ for } n = 1$$