New identity for Euler's Totient Function?
This is an interesting thing to notice, and you should be pleased.
As you guessed, and as Steven Stadnicki pointed out, this is not new; it follows quickly from two important properties of the $\phi$ function:
- $\phi(p^n) = (p-1)p^{n-1}$ when $p$ is prime
- $\phi(mn) = \phi(m)\phi(n)$ when $m$ and $n$ have no common factor
In particular, you suggested that your formula might be useful for calculating $\phi(n)$ for “large-ish $n$”, but observe that your formula requires knowing the radical of $n$, which is not in general any easier to find than the prime factorization of $n$. (And when the radical of $n$ is equal to $n$, as it is when $n$ is squarefree, your formula is no help at all.) But given $n = p_1^{a_1}\cdots p_k^{a_k}$ one has from (1) and (2) above that $$\begin{align} \phi(n) & = \phi\bigl(p_1^{a_1}\bigr)\cdots\phi\bigl(p_k^{a_k}\bigr) \\ & = (p_1-1)p_1^{a_1-1}\cdots (p_k-1)p_k^{a_k-1} \\ & = \frac{n}{p_1\cdots p_k}(p_1-1)\cdots(p_k-1) \\ & = n\left(\frac{p_1-1}{p_1}\right)\cdots \left(\frac{p_k-1}{p_k}\right) \tag{$\heartsuit$}\\ & = n\left(1-\frac1{p_1}\right)\cdots\left(1-\frac1{p_k}\right) \end{align}$$ which is well-known, and not significantly harder to compute (or perhaps easier) than your formula. The next-to last line ($\heartsuit$) is very similar to your formula.
Your identity can be written as $$ \frac{\phi(n)}{n}=\frac{\phi(\operatorname{rad}(n))}{\operatorname{rad}(n)} $$ In this form, it is obvious, because $$ \frac{\phi(m)}{m}=\prod_{p\mid m}\left(1-\frac1p\right) $$ and $n$ and $\operatorname{rad}(n)$ have exactly the same prime factors.
Still, a nice observation, well done!