So I firstly worked with the case where $n$ and $m$ are corpime and I found out that there is a nice solution using the Chinese remainder theorem.
Now I want to proof the following relation, this time for arbitrary numbers $$\phi(m \cdot n) = \frac{\gcd(m,n)\cdot\phi(m)\cdot\phi(n)}{\phi(\gcd(m,n))}.$$
However, the structure of the right-hand side is completely unclear to me so far and any hint would be highly appreciated.
You can use the fact that $\phi$ is multiplicative when $m,n$ are coprime to reduce your formula to the case where $m,n$ are powers of the same prime $p$. Just factor $m,n$ into primes, find $\phi$ of the product of the powers of the same prime, then multiply the results for each prime together. For example, if $m=p^aq^br^c, n=p^dq^es^f$ we have $\phi(m \cdot n)=\phi(p^{a+d}q^{b+e}r^cs^f)=\phi(p^{a+d})\phi(q^{b+e})\phi(r^c)\phi(s^f)$. Now let $m=p^a, n=p^b$ with $p$ prime and $a \ge b$. Then $$\mathbf{gcd}(m,n)=n=p^b\\ \phi(p^b)=(p-1)p^{b-1}\\\phi(m\cdot n)=\phi(p^{a+b})=(p-1)p^{a+b-1}\\ \frac{\mathbf{gcd}(m,n)\cdot\phi(m)\cdot\phi(n)}{\phi(\mathbf{gcd}(m,n))}=\frac{p^b(p-1)^2p^{a+b-2}}{(p-1)p^{b-1}}=(p-1)p^{a+b-1}$$
