First take it easy:
T(n) = T(n-n^(1/2)) + n, number of iteration is n^(1/2), in each iteration you'll have n-ksqrt(n) time complexity, so total time complexity is: ∑n-ksqrt(n) for 0<=k<=sqrt(n), which is n^(3/2).
Now solve your own problem:
T(n) = T(n^(1/2))+T(n-n^(1/2)) + n
again calculate number of steps till arriving to zero or 1: first part `T(n^(1/2)) takes O(log log n) time, and second part takes O(sqrt(n)) times to arriving to zero or 1 (See my answer to related question), So second part dominates first part, Also in each iteration time complexity of second part causes to sqrt(n) extra item, which doesn't effect to the first part time complexity (n-sqrt(n)), so your total runtime is n*sqrt(n).