$f(z)=xy+iy$
$f(z)=u(x,y)+iv(x,y)$
$u=xy$, $v=y$
$\frac{\partial u}{\partial x}=y$, $\frac{\partial v}{\partial x}=0$
$\frac{\partial u}{\partial y}=x$, $\frac{\partial v}{\partial y}=1$
We see that $\frac{\partial u}{\partial x} \ne \frac{\partial v}{\partial y}$, except for $y=1$
Also, $\frac{\partial u}{\partial y} \ne -\frac{\partial v}{\partial x}$, except for $x=0$
The Cauchy-Riemann equations are not satisfied anywhere in the z-plane except at the point $x=0$, $y=1$, i.e. $z=i$
$f^{'}(z)=\lim_{\delta z-> 0}\frac{f(z+\delta z)-f(z)}{\delta z}$
Best Answer
The Cauchy-Riemann equations must hold everywhere for a function to be analytic, and they fail here. Your computations are correct and show that the equations do not hold in any non-empty open subset of $\mathbb{C}$, so the function is not analytic anywhere.