The answer to both of your questionquestions is no.
Notice that $d(2, 3) = 1$. However, for any $\varepsilon > 0$, there exists a number $2 - \varepsilon < x < 2$ such that $d(x, 3) = 0$ and thus it is not a continuous function. To see this, take $n$ to be some large positive integer and $m$ the unique integer satisfying $$2^{m - 1} < 3^n < 2^m$$ or equivalently $$\left( 2^{1 - \frac{1}{m}} \right)^m < 3^n < 2^m$$ therefore there is a number $2^{1 - \frac{1}{m}} < x < 2^m$ such that $x^m = 3^n$ and therefore $d(x, 3) = 0$. Clearly, taking $n \to \infty$ we can let $x$ be arbitrarily close to $2$. In fact, this argument shows that for any $y > 1$, the set of $x > 1$ such that $d(x, y) = 0$ is dense in the set of numbers whichthat are larger than $1$.
As for the triangle inequality:, it is easy to see that $d(2, 3) = 1$ and $d(2, 16) = 0$. FurtherforeFurthermore, $d(3, 16) > 1$ because the only solutions to the equation $$2^m - 3^n = \pm 1$$ in positive integers $m, n$ are $(m, n) = (1, 1), (2, 1), (3, 2)$, which shows that there are no solutions to $$16^m - 3^n = \pm 1$$ and, thus, $d(3, 16) > d(3, 2) + d(2, 16)$.
