Verified answer

Let z = x + iy, and Δz = h + ik.

Then,

f '(z) = lim(Δz→0) [f(z + Δz) - f(z)] / Δz

.......= lim((h,k)→(0,0)) [f((x+h) + i(y+k)) - f(x + iy)] / (h + ik)

.......= lim((h,k)→(0,0)) [(y+k)/((x+h) - i(y+k)) - y/(x-iy)] / (h + ik), by definition of f

.......= lim((h,k)→(0,0)) [(y+k)(x-iy) - y((x+h) - i(y+k))] / [(x - iy) (h + ik) ((x+h) - i(y+k))]

.......= lim((h,k)→(0,0)) (kx - hy) / [(x - iy) (h + ik) ((x+h) - i(y+k))]

Now, let's try different paths to the origin.

(i) h = 0:

lim(k→0) kx / [(x - iy) * ik * (x - i(y+k))]

= lim(k→0) x / [i(x - iy)(x - i(y+k))]

= -ix / (x - iy)^2.

(ii) k = 0:

lim(h→0) -hy / [(x - iy) * h ((x+h) - iy)]

= lim(h→0) -y / [(x - iy) ((x+h) - iy)]

= -y/(x - iy)^2.

If f were differentiable at z, then we would need the limits in (i) and (ii) to be equal.

So, -ix / (x - iy)^2 = -y/(x - iy)^2

==> x = iy

==> x = y = 0, since x and y are real.

However, z = 0 + 0i = 0 is not part of the domain for f.

Hence, f is not differentiable for any z in C.

I hope this helps!