The directional derivative of f at p in the direction of any vector $\bar{Q}$ is $\bar{V}f\bigg(\frac{\bar{a}}{|a|}\bigg)$

$f = d = x^2 ycosz \\ \therefore \bar{V}d = \frac{\partial d}{\partial x}\hat{i} + \frac{\partial d}{\partial y}\hat{j} + \frac{\partial d}{\partial z}\hat{k} \\ \therefore \frac{\partial d}{\partial x} = 2xycosz$

Here

$\frac{\partial d}{\partial y} = x^2 cosz \\ \frac{\partial d}{\partial z} = -x^2 ysinz \\ \therefore \bar{V}d = (2xycosz)\hat{i} + (x^2 cosz)\hat{j} + (-x^2y sinz)\hat{k}$

Now,

$\bar{V}d|_{(1,2,\frac{\pi}{2})} = 0 + 0 - 1^2 * 2 * 1 \hat{k} \\ \bar{V}d|_{(1,2,\frac{\pi}{2})} = -2\hat{k}$

Now unit vector $\hat{a}$ in the direction towards $\bar{a} = 2\bar{i} + 3\bar{j} + 2\bar{k}$

$$\therefore \hat{a} = \frac{\bar{a}}{|a|} = \frac{2\bar{i} + 3\bar{j} + 2\bar{k}}{\sqrt{2^2 + 3^2 + 2^2}} = \frac{2\bar{i} + 3\bar{j} + 2\bar{k}}{\sqrt{17}}$$

Now required directional derivative is

$\bar{V}d_{(1,2,\frac{\pi}{2})}\hat{a} = -2\hat{k}\bigg(\frac{2\bar{i} + 3\bar{j} + 2\bar{k}}{\sqrt{17}}\bigg) = \frac{-4}{\sqrt{17}}$

Directional derivative is $\frac{-4}{\sqrt{17}}$