I following problem was given as a homework, I have explained how I approached it I need to know if it was correct and even then if it there wasn't any easier way, because that way only had tedious calculations. Find the points in $ \Bbb C $ such that $ f(z)= \frac{(\bar Z)^2}{Z}$ when $z $ in not equal to 0 and $f(z)=0 $ when $z=0$
1)f is differentiable 2) f is analytic 3) f satisfies CR equations.
I let $z=x+iy$ and then $f(z) = \frac{(x-iy)^3}{x^2+y^2}$
by simplifying it gave $u(x,y) =\frac{(x^3-3xy^2)}{ (x^2+y^2)} $ and $ v(x,y)=\frac{(y^3-3x^2y)}{ (x^2+y^2 )}$
where $f=u+iv$
By computing the partial derivatives and the equations are not statisfied at a non zero point. And I can show it is not differentiable at $0$ Therefore it is differentiable no where and analytic nowhere?
Fixing your typo in $v$, I get that $$\frac{\partial u}{\partial x} = 3x^2-3y^2 = -(-3x^2 + 3y^2) = -\frac{\partial v}{\partial y}$$
$$\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} = -6x y$$
So in a sense you get the opposite of the Cauchy-Riemann equations.
Indeed the only place the original equations are satisified is at $(0,0)$.