Answer:
The given function is differentiable at y = 1.
At y = 1, f'(z) = 0
Step-by-step explanation:
As per the given question,

Let z = x + i y
Suppose,


On computing the partial derivatives of u and v as:


And


According to the Cauchy-Riemann equations

Now,


Therefore,
holds only.
This means,
2y - 2 = 0
⇒ y = 1
Therefore f(z) has a chance of being differentiable only at y =1.
Now we can compute the derivative
![f'(z)=\frac{1}{2}[(u'_{x}+iv'_{x})-i(u'_{y}+iv'_{y})]](https://tex.z-dn.net/?f=f%27%28z%29%3D%5Cfrac%7B1%7D%7B2%7D%5B%28u%27_%7Bx%7D%2Biv%27_%7Bx%7D%29-i%28u%27_%7By%7D%2Biv%27_%7By%7D%29%5D)
![f'(z) =\frac{1}{2}[(2x+i(2-2y))-i(2y-2+i(-2x))]](https://tex.z-dn.net/?f=f%27%28z%29%20%3D%5Cfrac%7B1%7D%7B2%7D%5B%282x%2Bi%282-2y%29%29-i%282y-2%2Bi%28-2x%29%29%5D)

At y = 1
f'(z) = 0
Hence, the required derivative at y = 1 , f'(z) = 0