$\because f(x)=k x^3-9 x^2+9 x+3(k>0)$ is increasing for all $x$.
$\because f(x)$ is increasing, then $f^{\prime}(x) \geq 0$
$$
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x) \geq 0 \\
& \Rightarrow 3 k x^2-18 x+9 \geq 0 \\
& \Rightarrow \quad k x^2-6 x+3 \geq 0 \\
&
\end{aligned}
$$
Using concept of quadratic polynomial for $k>0$
It is possible only, when
$$
\begin{aligned}
& D \leq 0 \\
& b^2-4 a c \leq 0 \\
& \Rightarrow \quad(-6)^2-4(k)(3) \leq 0 \Rightarrow 36-12 k \leq 0 \\
& \Rightarrow \quad 12 k \geq 36 \Rightarrow k \geq 3 \\
&
\end{aligned}
$$
$\therefore$ Solution is $k \geq 3$.