Answer
$\operatorname{curl} \mathbf{H} = (x^2 + 2xz + x, \; 0, \; -z^2 - 2xz - z)$
Work Step by Step
\[
\mathbf{F} = (x, 0, -z), \quad \mathbf{G} = (x^2, y, z^2)
\]
\[
\mathbf{H} = \mathbf{F} \times \mathbf{G} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x & 0 & -z \\
x^2 & y & z^2
\end{vmatrix}
= \mathbf{i}(0 \cdot z^2 - (-z) \cdot y)
- \mathbf{j}(x \cdot z^2 - (-z) \cdot x^2)
+ \mathbf{k}(x \cdot y - 0 \cdot x^2)
= \mathbf{i}(yz) - \mathbf{j}(xz^2 + x^2 z) + \mathbf{k}(xy)
\]
\[
\operatorname{curl} \mathbf{H} = \nabla \times \mathbf{H} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
yz & -xz^2 - x^2 z & xy
\end{vmatrix}
\]
\[
\operatorname{curl} \mathbf{H} =
\mathbf{i} \Bigl( \frac{\partial}{\partial y}(xy) - \frac{\partial}{\partial z}(-xz^2 - x^2 z) \Bigr)
- \mathbf{j} \Bigl( \frac{\partial}{\partial x}(xy) - \frac{\partial}{\partial z}(yz) \Bigr)
+ \mathbf{k} \Bigl( \frac{\partial}{\partial x}(-xz^2 - x^2 z) - \frac{\partial}{\partial y}(yz) \Bigr)
\]
\[
\operatorname{curl} \mathbf{H} =
\mathbf{i} (x - (-2xz - x^2)) - \mathbf{j} (y - y) + \mathbf{k}((-z^2 - 2xz) - z)
= \mathbf{i} (x^2 + 2xz + x) - \mathbf{j} (0) + \mathbf{k}(-z^2 - 2xz - z)
\]
