jmc

Using the given $f(3x)=3f(x)$ repeatedly, we have that $$f(2001)=3f\left(\frac{2001}{3}\right)=3^{2}f\left(\frac{2001}{3^2}\right)=\cdots =3^{6}f\left(\frac{2001}{3^6}\right).$$Since $1\le 2001/3^6\le 3,$ we can apply the second part of the definition of $f$ to get $$f(2001)=3^6\left(1-\left|\frac{2001}{3^6}-2\right|\right)=3\cdot 3^6-2001=186.$$Therefore, we want the smallest $x$ for which $f(x)=186.$ Note that the range of $f(x)$ in the interval $x\in [1,3]$ is $[0,1].$ Since $f(3x)=3f(x)$ for all $x,$ it follows that the range of $f(x)$ in the interval $x\in [3,9]$ is $[0,3].$ Similarly, for each $k,$ the range of $f(x)$ in the interval $x\in [3^k,3^{k+1}]$ is $[0,3^k].$ Therefore, if $f(x)=186,$ then $3^k\ge 186,$ so $k\ge 5.$

We search the interval $x\in [3^5,3^6]=[243,729].$ We want $f(x)=186,$ and for any $x$ in this interval, we have $f(x)=3^{5}f\left(\frac{x}{3^5}\right).$ Therefore, letting $y=\frac{x}{3^5},$ we want $f(y)=\frac{186}{3^5}=\frac{186}{243},$ where $y\in [1,3].$ That is, $$1-|y-2|=\frac{186}{243}⟹|y-2|=\frac{57}{243}.$$The smaller of the two solutions to this equation is $y=2-\frac{57}{243}=\frac{429}{243}.$ Thus, $x=3^{5}y=\boxed{429}.$