Let z be a point of Z. Because f : Z --> X is finite, then O_Z,z is an O_X,f(z)-algebra of the same dimension. By assumption, O_Y,z is an O_Z,z-algebra of relative dimension d.

By the theory of regular sequences (see the refernce in the comments), the sequence f_i,z in O_Y,z, of stalks of the functions f_i, is regular if the relative dimension of O_Y,z over O_Z,z = O_Y,z/(f_i,z) is equal to d. This is the case because of the assumptions. Therefor I conclude that the sequence is regular. This means that the embedding Z --> Y is a regular embedding. This implies that f : Z --> X is flat. Q.E.D.

Let z be a point of Z. Because f : Z --> X is finite, then O_Z,z is an O_X,f(z)-algebra of the same dimension. By assumption, O_Y,z is an O_X,f(z)-algebra of relative dimension d.

By the theory of regular sequences (see the refernce in the comments), the sequence f_i,z in O_Y,z, of stalks of the functions f_i, is regular if the relative dimension of O_Y,z over O_Z,z = O_Y,z/(f_i,z) is equal to d. This is the case because of the assumptions. Therefor I conclude that the sequence is regular. This means that the embedding Z --> Y is a regular embedding. This implies that f : Z --> X is flat. Q.E.D.