The roots of equation $x3^x=1$ are

Hint:

$$3^x = \frac{1}{x}$$ Has only $1$ solution.

For a simple answer plot the graphs of $y_1=3^x$ and $y_2=\frac{1}{x}$, (that are elementary), and see that these graph intersects only at a point $x_0$ such that $0<x_0<1$ because:

1) $3^0=1$ and $ \frac{1}{x} \to +\infty$ for $x \to 0^+$

2) $y_1(1)=3^1=3>y_2(1)=\frac{1}{1}=1$

3) the two functions are continuous in $(0,1]$.

4) for $x>0$ $y_1$ is monotonic increasing and $y_2$ is monotonic decreasing and for $x<0$: $y_1>0$ and $y_2<0$.

If you want the value of $x_0$ this cannot be done with elementary functions. You can use the Lambert $W$ function that is defined as: $$ W(xe^x)=x $$ so, from $$ x3^x=1 \iff xe^{x\ln 3}=1 $$ using $x\ln 3=t$ we find: $$ te^t=\ln 3\quad \Rightarrow \quad t=W(\ln 3) $$ and $$ x_0=\frac{W(\ln 3)}{\ln 3} $$