= Solution
Use <prime factorization> and <multiplicativity> to check the identity one <prime> at a time. Fix $p$ and put $a=v_p(m)$, $b=v_p(n)$, $r=\min(a,b)$. The local factors to compare are
$$
\phi(p^{a+b})\phi(p^r)\quad\text{and}\quad
\phi(p^a)\phi(p^b)p^r.
$$
If either exponent is zero, $r=0$ and the equality follows from $\phi(1)=1$. If both are positive, the prime-power formula makes both sides
$$
p^{a+b+r-2}(p-1)^2.
$$
Multiplying the local identities over all <primes> and using $\gcd(m,n)=\prod_p p^{\min(v_p(m),v_p(n))}$ gives
$$
\boxed{\phi(mn)\phi(\gcd(m,n))=\phi(m)\phi(n)\gcd(m,n).}
$$
This accounts for repeated <prime factors> shared by the two inputs, which ordinary <coprime> <multiplicativity> alone would not remove.
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