Explanation

(a)
Given, $z = \tan{(y + ax) + \sqrt{y - ax}}$
$\Rightarrow z_{x} = \sec^{2}{(y + ax)a + \frac{1}{2\sqrt{y - ax}}( - a)}$
$\Rightarrow z_{xx} = 2\sec^{2}{(y + ax)\tan{(y + ax)a^{2} + \frac{1\left( - a^{2} \right)}{4(y - ax)^{3/2}}}}$
and $z_{y} = \sec^{2}{(y + ax) + \frac{1}{2\sqrt{y - ax}}}$
$\Rightarrow z_{yy} = 2\sec^{2}{(y + ax)\tan{(y + ax) - \frac{1}{4(y - ax)^{3/2}}}}\$
$\therefore z_{xx} - a^{2}z_{yy} = 0$