Recurence relation $D(n) \le D(\frac{2}{3}n) + 2$ We know that 
$D(1) = 0$.
Now from the inequality,
$$D(n) \le D\left(\frac{2}{3}n\right) + 2$$
it should follow that 
$$D(n) \le 2\cdot \log_{\frac{3}{2}}(n).$$
*Edit: the logarithm has a base $\frac{3}{2}$
How do we get that last bound?

(Disclaimer: the following assumes this is one of those contexts where certain hand-waving is allowed/expected, otherwise see the posted comments for why the problem is not well posed.)
Let $\,n = (3/2)^k \iff k = \log_{3/2}(n)\,$, and define $T(k)=D\big((3/2)^k\big)=D(n)\,$, then:
$$
T(k)=D \left(\left(\frac{3}{2}\right)^k \right) \;\le\; D \left(\frac{2}{3} \cdot \left(\frac{3}{2}\right)^k \right) + 2 = D \left(\left(\frac{3}{2}\right)^{k-1} \right) + 2 = T(k-1) + 2
$$
Next, by telescoping and using that $T(0)=D(1)=0\,$:
$$
T(k) \le T(k-1)+ 1 \cdot 2 \le T(k-2) + 2 \cdot 2 \le \cdots \le T(0)+k \cdot 2 = 2k
$$
Rewriting the latter in terms of $n\,$, it follows that $D(n) = T(k) \le 2 k = 2 \log_{3/2} n$.
