How to solve this linear congruence $46x\equiv 6\pmod {202}$

As already noticed, $$46x\equiv6\mod 2002\iff 23x\equiv 3\mod 101.$$ There remains to find the inverse of $23$ modulo $101$. This is obtained with the Extended Euclidean algorithm: \begin{array}{rrrl} \hline r_i&u_i&v_i&q_i\\\hline101&0&1\\23&1&0&4\\\hline 9&-4&1&2\\5&9&-2&1\\4&-13&3&1\\1&22&-5\\\hline \end{array} The last row yields the coefficients of a Bézout's relation between $13$ and $101$: $$22\times 23-5\times101=1,$$ whence the inverse of $23$ mod $101$: $22$, and the solution of the congruence: $$x\equiv 22\times3=66\mod 101.$$

You must cancel $\,2\,$ from the modulus too, as is clear by converting to equation form, i.e.

$\ 46x\equiv 6\pmod{\!202}\!\iff\! 46x\!-\!202y=6\!\!\!\overset{\,\large \div\, 2}\iff\! 23x\!-\!101y=3\!\iff\! \color{#c00}{23x\equiv 3}\pmod{\!101}$

Finally below we give a couple convenient ways to compute the fraction $\, \color{#c00}{x\equiv 3/23}\pmod{101}$

${\rm mod}\,\ 101\!:\ \ \color{#c00}{\dfrac{3}{23}}\!\!\!\overset{\ \ \ \times 4_{\phantom{._{._.}}}}\equiv\!\!\dfrac{12}{92}\equiv\,\dfrac{12}{-9}\,\equiv\, \dfrac{4}{-3}\equiv\dfrac{105}{-3}\equiv\color{#0a0}{-35}\ $ by Gauss's algorithm. $ $ Alternatively

${\rm mod}\,\ 101\!: \ \dfrac{0}{101} \overset{\large\frown}\equiv \underbrace{\color{#c00}{\dfrac{3}{23}}\overset{\large\frown}\equiv \dfrac{-12}9\equiv \color{#90f}{\dfrac{-4}3}\overset{\large\frown}\equiv \color{#0a0}{\dfrac{35}{-1}}} _{\Large \begin{align}\color{#c00}3\ \ \ \ \ -\ \ \ \ \ &8(\color{#90f}{-4})\ \ \ \ =\ \ \ \, \color{#0a0}{35}\\ \color{#c00}{23}\ \ \ \ -\ \ \ \ \ &8\ (\ \color{#90f}3\ )\ \ \ \ =\ \ \, \color{#0a0}{-1}\end{align}}\ $ by the fractional extended Euclidean algorithm

where the underbace shows the basic reduction step, i.e. perform Euclid's algorithm to reduce the denominator sequence, doing the same in parallel to the numerators (explained in the prior link). Above we also used an intermediate step (cancelling $\,3\,$ from $-12/9\,$ to get $\color{#90f}{-4/3})\,$ before applying the underlined Euclidean reduction step (cancelling $\,3\,$ is valid because $3$ is coprime to $101).$

Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.