Answer
$\text{curl } \mathbf{F} = \frac{2x}{x^2 + y^2} \mathbf{k} $
Work Step by Step
$F(x, y, z) = \arctan\left(\frac{x}{y}\right)\mathbf{i} + \ln\sqrt{x^2 + y^2},\mathbf{j} + \mathbf{k}$
$\text{curl } F = \nabla \times F = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R
\end{vmatrix}$
where $P = \arctan\left(\frac{x}{y}\right)$, $Q = \ln\sqrt{x^2 + y^2}$, and $R = 1$.
$\text{curl } F = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\arctan\left(\frac{x}{y}\right) & \ln\sqrt{x^2 + y^2} & 1
\end{vmatrix}$
$\text{curl } \mathbf{F} = \left( \frac{\partial}{\partial y}(1) - \frac{\partial}{\partial z}(Q) \right)\mathbf{i} - \left( \frac{\partial}{\partial x}(1) - \frac{\partial}{\partial z}(P) \right)\mathbf{j} + \left( \frac{\partial}{\partial x}(Q) - \frac{\partial}{\partial y}(P) \right)\mathbf{k}$
$= \left( 0 - 0 \right)\mathbf{i} - \left( 0 - 0 \right)\mathbf{j} + \left( \frac{\partial}{\partial x}(\ln\sqrt{x^2 + y^2}) - \frac{\partial}{\partial y}(\arctan\left(\frac{x}{y}\right)) \right)\mathbf{k}$
$= \left( 0 \right)\mathbf{i} - \left( 0 \right)\mathbf{j} + \left( \frac{x}{x^2 + y^2} - \left( -\frac{x}{x^2 + y^2} \right) \right)\mathbf{k}$
$= \left( 0 \right)\mathbf{i} - \left( 0 \right)\mathbf{j} + \left( \frac{2x}{x^2 + y^2} \right)\mathbf{k}$
