If $f(z)=z|z-1|^2$ where $z=x+iy$. I need to show where it is differentiable, and then from first principles find its derivative at each point.
I have started by saying $f=(x+iy)((x-1)^2+y^2)$
After doing Cauchy-Riemann, I have the following
$U_x=3x^2 -2x^2+x+xy^2\\ V_x = 2xy-2y\\ U_y= 2xy \\ V_y= x^2-2x+1+3y^2$
I am really stuck as to what to do next.
Solve the equations: $$ \begin{cases} U'_x = V'_y \\ U'_y = -V'_x \end{cases} $$
Your $U'_x$ seems wrong though. I get $$ \begin{cases} U'_x = 3x^2 -4x + y^2 +1 \\ U'_y = 2xy \\ V'_x = 2xy - 2y \\ V'_y = x^2-2x+1 + 3y^2 \end{cases} $$ Looking at $U'_y = -V'_x$ we get $2xy = -2xy+2y$, i.e. $4xy-2y = 0$ or $2y(2x-1) = 0$. Hence, $y=0$ or $x=1/2$.
If $y=0$, $U'_x = V'_y$ reduces to $$ 3x^2-4x + 1 = x^2-2x+1 \quad\iff\quad 2x(x-1) = 0 $$ i.e. $x=0$ or $x=1$.
On the other hand, if $x=1/2$ then $$ \frac34 - 2 + y^2 + 1= \frac14 - 1 + 1 + 3y^2 $$ which has no real solutions.
Since $U$ and $V$ are $C^1$, we see that your function is complex differentiable at $z=0$ and $z=1$ and no other points.
If you know the formalism of Wirtinger derivatives, the above computation will be a lot easier.