I didn't find any references yet, but it looks very interesting. Instead of doing the usual inclusion/exclusion on composite numbers, it does inclusion/exclusion on prime powers.
$$\pi(n) = -\sum\limits_{k=1}^{\log_2(n)}\mu(k)\sum\limits_{l=2}^{\lfloor \sqrt[k]{n} \rfloor} \left\lfloor \frac{\sqrt[k]{n}}{l}\right\rfloor \mu(l)\Omega(l)$$
It starts with $k=1$, where we get the number of primes and primes powers less than $n$ or in other word: $\pi(n)+\pi(\sqrt[2]{n})+\pi(\sqrt[3]{n})+\pi(\sqrt[4]{n})+\pi(\sqrt[5]{n})+...+\pi(\sqrt[\log_2(n)]{n})$
With $k=2$ we start to remove power of primes that are multiple of $2$ ($2^2,2^4,2^6,...,3^2,3^4,3^6,...$)
With $k=3$ we start to remove power of primes that are multiple of $3$ ($2^3,2^6,...,3^3,3^6,...$)
With $k=6$ we start to add the power of primes that are multiple of $6$ and were removed twice above ($2^6,...,,3^6,...$)
In the end we are left with $\pi(n)$
Now, to show it, I think you can use some properties of the binomial coeficient.
Let's look at $k=1$ again:
$$-\sum\limits_{l=2}^{n} \left\lfloor \frac{n}{l}\right\rfloor \mu(l)\Omega(l)$$
without $\Omega(l)$ we have the classic inclusion/exclusion which remove composites counted multiple times:
$$-\sum\limits_{l=2}^{n} \left\lfloor \frac{n}{l}\right\rfloor \mu(l)=\sum\limits_{p_i<=n}\lfloor \frac{n}{p_i} \rfloor-\Sigma\Sigma\lfloor \frac{n}{p_i\cdot p_j} \rfloor+\Sigma\Sigma\Sigma\lfloor \frac{n}{p_i\cdot p_j\cdot p_k} \rfloor-...=(n-1)$$
e.g. $210=2\cdot3\cdot5\cdot7$ is counted in $\lfloor \frac{n}{2} \rfloor$, $\lfloor \frac{n}{3} \rfloor$, $\lfloor \frac{n}{5} \rfloor$, $\lfloor \frac{n}{7} \rfloor$, what is counted twice is removed with $\lfloor \frac{n}{2\cdot3} \rfloor$, $\lfloor \frac{n}{2\cdot5} \rfloor$,$\lfloor \frac{n}{2\cdot7} \rfloor$,$\lfloor \frac{n}{3\cdot5} \rfloor$,$\lfloor \frac{n}{3\cdot7} \rfloor$,$\lfloor \frac{n}{5\cdot7} \rfloor$, what was removed too much is added back in $\lfloor \frac{n}{2\cdot3\cdot5} \rfloor$,$\lfloor \frac{n}{2\cdot3\cdot7} \rfloor$,$\lfloor \frac{n}{2\cdot5\cdot7} \rfloor$, $\lfloor \frac{n}{3\cdot5\cdot7} \rfloor$, and finaly $\lfloor \frac{n}{2\cdot3\cdot5\cdot7} \rfloor$ is removed.
In other word, with composites appearing multiple times (here having 4 distinct prime factors), we only count one of them in the end:
$$\binom{4}{1}-\binom{4}{2}+\binom{4}{3}-\binom{4}{4}=1$$
And this is a property of the binomial coeficients (here with $m$ distinct prime factors):
$$\sum\limits_{i=1}^{m}(-1)^{i+1}\binom{m}{i}=1$$
Now if you put back $\Omega(l)$, we have this:
$$-\sum\limits_{l=2}^{n} \left\lfloor \frac{n}{l}\right\rfloor \mu(l)\Omega(l)=1\cdot\sum\limits_{p_i<=n}\lfloor \frac{n}{p_i} \rfloor-2\cdot\Sigma\Sigma\lfloor \frac{n}{p_i\cdot p_j} \rfloor+3\cdot\Sigma\Sigma\Sigma\lfloor \frac{n}{p_i\cdot p_j\cdot p_k} \rfloor-4\cdot...$$
Now what happens to the composite of our exemple above is this:
$$1\cdot\binom{4}{1}-2\cdot\binom{4}{2}+3\cdot\binom{4}{3}-4\cdot\binom{4}{4}=0$$
And this is another property of the binomial coeficients:
$$\sum\limits_{i=1}^{m>1}(-1)^{i+1}\cdot i\cdot\binom{m}{i}=0$$
Note: for $m=1$ the above equation is equal to $1$.
What it means is that no composite are counted. Only numbers with 1 factor are (primes and prime powers).
For $k>1$ the resoning is probably the same, but I had not much time yet.
EDIT: Sorry for the late update, I couldn't look at it earlier. I guess that you already looked at it by now, but for thoose who didn't: It is indeed the same reasoning. for $k=2$, primes and prime powers are counted up to $\sqrt[2]{n}$ which is a count of prime+prime powers squared, for $k=3$, primes+primes power cubes are counted, and so on....
Best Answer
You can improve the $O(n^{3/4})$ algorithm by a constant factor of roughly three by computing only the odd summands of $k$ by using the identity:
$$\begin{array}{lll} f \left( n \right) & = & g \left( n \right) - g \left( \frac{n}{2} \right) - \sum_{3 \leq k \leq n, k \:\mathrm{odd}} f \left( \left\lfloor \frac{n}{k} \right\rfloor \right) \end{array}$$
which can be derived by subtracting $g(\frac{n}{2})$ from $g(n)$. The odd-only approach is hinted at in [Lehman 1960] (search for "in practice").
You can further improve the time complexity to $O(n^{2/3+\epsilon})$ if you can compute $f(k)$ for each $k\in \left\{1,..,u \right\}$ in $O(u^{1+\epsilon})$, as is typical for arithmetic functions by using a (possibly segmented) sieve.
Two references for the $O(n^{2/3})$ approach are [Pawlewicz 2008] and [Deleglise 1996].