help me clear a doubt on vector calculus I found that the explanation to option (d) is wrong as curl of vector $F$ is not zero hence not irrotational. What say?

 A: You are wrong. On finding curl F using determinat we got 0. Hence irrotational.
Curl F = $\left(- \frac{2x^3}{z^3} + \frac{2x^3}{z^3} \right) \hat{i} - \left(\frac{-6x^2y}{z^3} - \frac{-6x^2y}{z^3} \right) \hat{j} + \left(8y + \frac{3x^2}{z^2} - 8y - \frac{3x^2}{z^2} \right) \hat{k}$ 
= 0
A: You probably made a mistake in your calculations.
$$\mbox{curl}\, \vec F = \begin{vmatrix}
\hat i & \hat j & \hat k \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
4y^2+\frac{3x^2y}{z^2} & 8xy+\frac{x^3}{z^2} & 11-\frac{2x^3y}{z^3}
\end{vmatrix} \\[38pt]
= \left( \frac{\partial}{\partial y} \left(11-\frac{2x^3y}{z^3} \right) -  \frac{\partial}{\partial z} \left( 8xy+\frac{x^3}{z^2} \right)\right)\hat i \\[18pt] + 
\left( \frac{\partial}{\partial z} \left( 4y^2+\frac{3x^2y}{z^2} \right) -  \frac{\partial}{\partial x} \left( 11-\frac{2x^3y}{z^3} \right)\right) \hat j \\[18pt] +
\left( \frac{\partial}{\partial x} \left( 8xy+\frac{x^3}{z^2} \right) -  \frac{\partial}{\partial y} \left( 4y^2+\frac{3x^2y}{z^2} \right) \right) \hat k \\[18pt]
= \cdots = 0.
$$
