f(1) = {b,c,e,f}
f(3) = {d}
Do you have the same number of layers of curly brackets in your value for f(3)?
Notice that P = {f(1), f(2), f(3)} = {{b,c,e,f}, {a,g}, {d}}.
P obviously doesn't contain the empty set, so that partition property holds.
The elements of P don't overlap, so that partition property is also ok.
However, none of the elements of P contains the node X. So P doesn't cover all the elements of V. So that partition property isn't satisfied.
So P isn't a partition of V.