Answer
$\nabla \times (\mathbf{F} \times \mathbf{G}) = 9x \mathbf{j} - 2y \mathbf{k}$
Work Step by Step
\[
\mathbf{F} = (1, 3x, 2y), \quad \mathbf{G} = (x, -y, z)
\]
\[
\mathbf{H} = \mathbf{F} \times \mathbf{G} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 3x & 2y \\
x & -y & z
\end{vmatrix}
\]
\[
= \mathbf{i}(3xz + 2y^2) - \mathbf{j}(z - 2xy) + \mathbf{k}(-y - 3x^2)
\]
\[
\text{curl } \mathbf{H} = \nabla \times \mathbf{H} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
3xz + 2y^2 & -(z - 2xy) & -y - 3x^2
\end{vmatrix}
\]
\[
= \mathbf{i} \left( \frac{\partial}{\partial y} (-y - 3x^2) - \frac{\partial}{\partial z} (-(z - 2xy)) \right)
- \mathbf{j} \left( \frac{\partial}{\partial x} (-y - 3x^2) - \frac{\partial}{\partial z} (3xz + 2y^2) \right)
+ \mathbf{k} \left( \frac{\partial}{\partial x} (-(z - 2xy)) - \frac{\partial}{\partial y} (3xz + 2y^2) \right)
\]
\[
= \mathbf{i} ( -1 - (-1) ) - \mathbf{j} ( -6x - 3x ) + \mathbf{k} ( 2y - 4y )
\]
\[
= 0\mathbf{i} + 9x \mathbf{j} - 2y \mathbf{k}
\]
