State Cauchy-Riemann equations. Show that f(z) x*+ iy

Step 1
Cauchy-Riemann Equations:
A necessary condition that the function f=u+iv is differentiable at a point ${\displaystyle z_0=x_0+i{y}_0}$ is that the partial derivatives ${\displaystyle u_x,u_y,v_x,v_y}$ exists and ${\displaystyle u_x=v_y,u_y=-v_x}$ at the point ${\displaystyle (x_0,y_0)}$
Step 2
Given equation is f=x+iy
Step 3
Let, f(z)=u(x,y)+iv(x,y)
Comparing,
u(x,y)=x
v(x,y)=y
Step 4
Then,
${\displaystyle u_x=1}$
${\displaystyle u_y=0}$
${\displaystyle v_x=0}$
${\displaystyle v_y=1}$
So, ${\displaystyle u_x=v_y\text{and}{u}_y=-v_x}$ at (0,0).
So, Cauchy-Riemann equations are satisfied at the origin.
Step 5
But the Cauchy-Riemann equations are satisfied only at the point z=0.
Hence, f(z)=x+iy can not have derivative at any point ${\displaystyle z\ne 0}$.
So, the given function is not analytic.