The Cauchy-Riemann equations are
\begin{align}
\dfrac{\partial u}{\partial x} & = \dfrac{\partial v}{\partial y}\\
\dfrac{\partial v}{\partial x} &= -\dfrac{\partial u}{\partial y}
\end{align}
In your case, $u(x,y) = x + \sqrt{x^2+y^2}$ and $v(x,y) = y$. Assuming $(x,y) \neq (0,0)$, the partial derivatives are
\begin{align}
\dfrac{\partial u}{\partial x} & = 1 + \dfrac{x}{\sqrt{x^2+y^2}}\\
\dfrac{\partial v}{\partial x} & = 0\\
\dfrac{\partial u}{\partial y} & = \dfrac{y}{\sqrt{x^2+y^2}}\\
\dfrac{\partial v}{\partial y} & = 1
\end{align}
Hence, from the Cauchy-Riemann equations, we get that
$$1 + \dfrac{x}{\sqrt{x^2+y^2}} = 1 \implies \dfrac{x}{\sqrt{x^2+y^2}} = 0$$
$$\dfrac{y}{\sqrt{x^2+y^2}} = 0$$
This has no solutions since $(x,y) \neq (0,0)$. Hence, the function is not differentiable on $\mathbb{C} \backslash \{(0,0)\}$. The only point we need to check whether it is differentiable is $(0,0)$. At this point, we can check for differentiability directly from the definition. You will find that it is also not differentiable at $(0,0)$. Hence, the function is nowhere analytic.