Given that $f(x)=k x^3-9 k x^2+9 x+3$ is an increasing function on R set of real numbers, then $k$ belongs to

The function $f(x) = kx^3 - 9kx^2 + 9x + 3$ is increasing on $\mathbb{R}$ if and only if its derivative $f'(x) \ge 0$ for all $x \in \mathbb{R}$.

Calculating the derivative:
$f'(x) = 3kx^2 - 18kx + 9$.

For $f'(x) \ge 0$ for all $x$, the quadratic expression $3kx^2 - 18kx + 9$ must satisfy two conditions:
1. The coefficient of $x^2$ must be positive, so $3k > 0$, which implies $k > 0$.
2. The discriminant $D$ must be less than or equal to zero, so $D \le 0$.

Calculating the discriminant $D = b^2 - 4ac$:
$D = (-18k)^2 - 4(3k)(9) = 324k^2 - 108k$.

Setting $D \le 0$:
$324k^2 - 108k \le 0$
$108k(3k - 1) \le 0$.

Since $k > 0$, we divide by $108k$ to get:
$3k - 1 \le 0$
$3k \le 1$
$k \le \dfrac{1}{3}$.

Combining $k > 0$ and $k \le \dfrac{1}{3}$, we get $k \in \left(0, \dfrac{1}{3}\right]$.

Answer: $\left(0, \dfrac{1}{3}\right]$