Since you seem to be a bit lost. read https://en.wikipedia.org/wiki/De_Morgan's_laws#Engineering
let me show one way. Often we use X or Y or Z for outputs or f(ABCDEFGH)=...
\$Y=((A\cdot B)\cdot (C\cdot D))\cdot ((E\cdot F)\cdot (G\cdot H))\$
Since you must solve using 2in-NAND gates , I will use ! to indicate an inverted logic. (sometimes you will find ! used before or after brackets, but you must be consistent!)
AND = \$(A\cdot B)!! = (A\cdot B) \$ with both inputs of a NAND gate joined to make it an inverter (INV).
\$Y=(~~(A\cdot B)!!\cdot (C\cdot D)!!~~ )!!\cdot ((E\cdot F)!!\cdot (G\cdot H)!!)!!\$
Cleaner notation removes the dot for AND but leaves + for OR.
\$Y=~((AB~!!)(CD)!! ) ~ ((EF)!!(GH)!!)!!\$
I should have labelled the inputs with ABC etc but I was too lazy.
Here is what it looks like with my simulator that denotes H,L for 1,0 where you can carefully click on any input ( without disconnecting it by dragging the mouse click ;)
So you see output is H only when all inputs are H.