Finding Mixed Nash Equilibria in a $3\times 3$ Game
Assume the pure strategy $s_i$ is strictly dominated by $s_i'$. This means that $s_i$ is not a best response to any of the opponent's pure strategies. For whatever strategy the other player is playing, the strategy $s_i'$ would yield a larger payoff. Even if $s_i'$ is a mixed strategy, for any strategy the opponent is playing you can select a pure component of $s_i'$ with a larger payoff than $s_i$ as the payoff of $s_i'$ is the average of the payoffs of its pure components.

A reversal of this is that if a pure strategy is a best response to any pure strategy of the opponent, it cannot be strictly dominated.

From the "underline the best response" method you will see that all the strategies of all the players are best response to at least one strategy of the opponent, hence none of them are strictly dominated.

Unfortunately, this still does not tell you much about the mixed equilibria. With different supports it is possible to have more than one.



An example:

In the game
$$
\begin{array}
\\&amp;L&amp;C&amp;R\\
T&amp;1,1&amp;0,0&amp;0,0\\
M&amp;0,0&amp;1,1&amp;0,0\\
B&amp;0,0&amp;0,0&amp;1,1
\end{array}
$$
the strategy profile where all strategies are mixed with probability $1/3$ is a mixed equilibrium, but the strategy profile where the Row player mixes $T$, $M$ while the Column player mixes $L$ and $C$ with probability $1/2$ is also a mixed equilibrium. (There are two other mixed equilibria and three pure ones.)



So there is no trivial method to find all the mixed equilibria, one usually has to use a trick or work through the different cases for all the supports.

Wilson's oddness theorem tells you that there are an odd number of equilibria, except in special "degenerate" cases.