Construct a counterexample to show that 𝑧²/(𝑧̅)² does not have a limit as z --> 0.?

Not clear what sort of 'counter example' is required. See solution to Q2 here: http://www.math.ualberta.ca/~xichen/math31110f/hw3sol.pdf

Hints: Compute (expand) (x+yi)². Ditto for (𝑧̅)². Show ur answers here.
Then we will take it from there.

Put z=x+iy then z²/z̅² = (x²−y²)/(x²+y²) + 2ixy/(x²+y²)

If z→0 along the line y=x then at any point on the path z²/̅z̅² = (x²−x²))/(x²+x²) + 2ix²/(x²+x²) = i
So along this path lim (z→0) z²/z̅² = i
If z→0 along the line y=2x then z²/̅z̅² = (3+4i)/5 so along this path lim (z→0) z²/z̅² = (3+4i)/5

The point is that as z→0, z²/̅z̅² approaches a value which depends on the path that z takes to 0.
For lim (z→0) z²/z̅² to exist z²/z̅² would have to approach the same value no matter what the path.