On the molybdenum(III) oxide-catalysed oxidation of alkanes to carbonyls
The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow.  Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$.
On industrial scale, formaldehyde is prepared from methanol by an  heterogeneous redox-reaction.  The oxygen needed may come from oxygen gas, or simply air.  Thus, the last step known as Formox process is
$$ \ce{2 CH3OH + O2 -&gt; 2 CH2O + 2 H2O} $$
(from wikipedia)
$\ce{Mo2O3}$ is deployed ease the advancement of the reaction, i.e., to perform the wanted reaction at lower temperature (heating costs money) than without this help, and to further increase rate or / and yield of said reaction compared performing this conversion at room temperature.