#CH_"3"-(CO)-O-CH_"3"-(CO)#. IN THIS, break the molecule from -O-(excluding -O- from both), and add -OH at both terminals. now write the name of the carboxylic acid so formed( ex- here ethanoic acid). remove the word 'acid' from the name and add 'anhydride' to get the answer.
on breaking form -O- you get two molecules of acid. if both are same then write the name only once, as I've written. if you get two different acids than write both in alphabetical order,
ex- #(CH_"3"-CH_"2"-CO-)O(-CO-CH_"3")#, on breaking this and following above rule you get 'propanoic acid' and 'ethanoic acid', so the given anhydride is' ethanoic propanic anhydride'