Answer
$\nabla \varphi = \frac{{{{\bf{e}}_r}}}{r}$
Work Step by Step
We have $\varphi = \ln r$, where $r = \sqrt {{x^2} + {y^2}} $.
The gradient of $\varphi $ is $\nabla \varphi = \left( {\frac{{\partial \varphi }}{{\partial x}},\frac{{\partial \varphi }}{{\partial y}}} \right)$.
Evaluate $\frac{{\partial \varphi }}{{\partial x}}$
$\frac{{\partial \varphi }}{{\partial x}} = \frac{1}{r}\frac{{\partial r}}{{\partial x}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\frac{\partial }{{\partial x}}\sqrt {{x^2} + {y^2}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\left( {\frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$
$\frac{{\partial \varphi }}{{\partial x}} = \frac{x}{{{r^2}}}$
Evaluate $\frac{{\partial \varphi }}{{\partial y}}$
$\frac{{\partial \varphi }}{{\partial y}} = \frac{1}{r}\frac{{\partial r}}{{\partial y}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\frac{\partial }{{\partial y}}\sqrt {{x^2} + {y^2}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\left( {\frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)$
$\frac{{\partial \varphi }}{{\partial y}} = \frac{y}{{{r^2}}}$
Therefore, $\nabla \varphi = \left( {\frac{{\partial \varphi }}{{\partial x}},\frac{{\partial \varphi }}{{\partial y}}} \right) = \frac{1}{r}\left( {\frac{x}{r},\frac{y}{r}} \right)$.
Since ${{\bf{e}}_r} = \left( {\frac{x}{r},\frac{y}{r}} \right)$, so $\nabla \varphi = \frac{{{{\bf{e}}_r}}}{r}$.
