1. **State the problem:** We are given a function involving $\tan(y+ax)$ and $(y-ax)^{3/2}$ and asked to find the value of $$\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2}$$ where $$z = \tan(y+ax) + (y-ax)^{3/2}.$$\n\n2. **Write the function:** $$z = \tan(y+ax) + (y-ax)^{3/2}.$$\n\n3. **Find the first partial derivatives:**\n- For $\frac{\partial z}{\partial x}$:\n - Use chain rule: $$\frac{\partial}{\partial x} \tan(y+ax) = \sec^2(y+ax) \cdot a.$$\n - For $(y-ax)^{3/2}$, treat $y$ as constant: $$\frac{\partial}{\partial x} (y-ax)^{3/2} = \frac{3}{2} (y-ax)^{1/2} \cdot (-a) = -\frac{3a}{2} (y-ax)^{1/2}.$$\nThus, $$\frac{\partial z}{\partial x} = a \sec^2(y+ax) - \frac{3a}{2} (y-ax)^{1/2}.$$\n\n- For $\frac{\partial z}{\partial y}$:\n - $$\frac{\partial}{\partial y} \tan(y+ax) = \sec^2(y+ax) \cdot 1 = \sec^2(y+ax).$$\n - $$\frac{\partial}{\partial y} (y-ax)^{3/2} = \frac{3}{2} (y-ax)^{1/2} \cdot 1 = \frac{3}{2} (y-ax)^{1/2}.$$\nThus, $$\frac{\partial z}{\partial y} = \sec^2(y+ax) + \frac{3}{2} (y-ax)^{1/2}.$$\n\n4. **Find the second partial derivatives:**\n- For $\frac{\partial^2 z}{\partial x^2}$:\n - Differentiate $\frac{\partial z}{\partial x}$ w.r.t. $x$:\n $$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left(a \sec^2(y+ax) - \frac{3a}{2} (y-ax)^{1/2} \right).$$\n - First term:\n $$a \cdot \frac{\partial}{\partial x} \sec^2(y+ax) = a \cdot 2 \sec^2(y+ax) \tan(y+ax) \cdot a = 2a^2 \sec^2(y+ax) \tan(y+ax).$$\n - Second term:\n $$-\frac{3a}{2} \cdot \frac{\partial}{\partial x} (y-ax)^{1/2} = -\frac{3a}{2} \cdot \frac{1}{2} (y-ax)^{-1/2} \cdot (-a) = \frac{3a^2}{4} (y-ax)^{-1/2}.$$\n So, $$\frac{\partial^2 z}{\partial x^2} = 2a^2 \sec^2(y+ax) \tan(y+ax) + \frac{3a^2}{4} (y-ax)^{-1/2}.$$\n\n- For $\frac{\partial^2 z}{\partial y^2}$:\n - Differentiate $\frac{\partial z}{\partial y}$ w.r.t. $y$:\n $$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left( \sec^2(y+ax) + \frac{3}{2} (y-ax)^{1/2} \right).$$\n - First term:\n $$2 \sec^2(y+ax) \tan(y+ax) \cdot 1 = 2 \sec^2(y+ax) \tan(y+ax).$$\n - Second term:\n $$\frac{3}{2} \cdot \frac{1}{2} (y-ax)^{-1/2} = \frac{3}{4} (y-ax)^{-1/2}.$$\n So, $$\frac{\partial^2 z}{\partial y^2} = 2 \sec^2(y+ax) \tan(y+ax) + \frac{3}{4} (y-ax)^{-1/2}.$$\n\n5. **Calculate the expression:**\n$$\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2} = \left(2a^2 \sec^2(y+ax) \tan(y+ax) + \frac{3a^2}{4} (y-ax)^{-1/2} \right) - a^2 \left(2 \sec^2(y+ax) \tan(y+ax) + \frac{3}{4} (y-ax)^{-1/2} \right).$$\n\n6. **Simplify:**\nDistribute $a^2$ in the second term:\n$$= 2a^2 \sec^2(y+ax) \tan(y+ax) + \frac{3a^2}{4} (y-ax)^{-1/2} - 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{3a^2}{4} (y-ax)^{-1/2} = 0.$$\n\n**Final answer:** $$\boxed{0}.$$