1. **Problem Statement:** Given the function $$z = \tan(y + ax) + (y - ax)^{3/2},$$ find the value of $$\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2}.$$\n\n2. **Step 1: Compute first partial derivatives.**\n- Let $$u = y + ax$$ and $$v = y - ax.$$\n- Then, $$z = \tan(u) + v^{3/2}.$$\n\nCalculate $$\frac{\partial z}{\partial x}$$:\n$$\frac{\partial z}{\partial x} = \sec^2(u) \frac{\partial u}{\partial x} + \frac{3}{2} v^{1/2} \frac{\partial v}{\partial x}.$$\nSince $$\frac{\partial u}{\partial x} = a$$ and $$\frac{\partial v}{\partial x} = -a,$$\nwe get\n$$\frac{\partial z}{\partial x} = a \sec^2(y + ax) - \frac{3a}{2} (y - ax)^{1/2}.$$\n\n3. **Step 2: Compute second partial derivative with respect to x.**\n$$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left(a \sec^2(u) - \frac{3a}{2} v^{1/2} \right) = a \frac{\partial}{\partial x} \sec^2(u) - \frac{3a}{2} \frac{\partial}{\partial x} v^{1/2}.$$\n\nCalculate each term:\n- $$\frac{\partial}{\partial x} \sec^2(u) = 2 \sec^2(u) \tan(u) \frac{\partial u}{\partial x} = 2a \sec^2(y + ax) \tan(y + ax).$$\n- $$\frac{\partial}{\partial x} v^{1/2} = \frac{1}{2} v^{-1/2} \frac{\partial v}{\partial x} = \frac{1}{2} (y - ax)^{-1/2} (-a) = -\frac{a}{2} (y - ax)^{-1/2}.$$\n\nTherefore,\n$$\frac{\partial^2 z}{\partial x^2} = a (2a \sec^2(y + ax) \tan(y + ax)) - \frac{3a}{2} \left(-\frac{a}{2} (y - ax)^{-1/2} \right) = 2a^2 \sec^2(y + ax) \tan(y + ax) + \frac{3a^2}{4} (y - ax)^{-1/2}.$$\n\n4. **Step 3: Compute first partial derivatives with respect to y.**\n$$\frac{\partial z}{\partial y} = \sec^2(u) \frac{\partial u}{\partial y} + \frac{3}{2} v^{1/2} \frac{\partial v}{\partial y}.$$\nSince $$\frac{\partial u}{\partial y} = 1$$ and $$\frac{\partial v}{\partial y} = 1,$$\nwe get\n$$\frac{\partial z}{\partial y} = \sec^2(y + ax) + \frac{3}{2} (y - ax)^{1/2}.$$\n\n5. **Step 4: Compute second partial derivative with respect to y.**\n$$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left( \sec^2(y + ax) + \frac{3}{2} (y - ax)^{1/2} \right) = \frac{\partial}{\partial y} \sec^2(u) + \frac{3}{2} \frac{\partial}{\partial y} v^{1/2}.$$\n\nCalculate each term:\n- $$\frac{\partial}{\partial y} \sec^2(u) = 2 \sec^2(u) \tan(u) \frac{\partial u}{\partial y} = 2 \sec^2(y + ax) \tan(y + ax).$$\n- $$\frac{\partial}{\partial y} v^{1/2} = \frac{1}{2} v^{-1/2} \frac{\partial v}{\partial y} = \frac{1}{2} (y - ax)^{-1/2} (1) = \frac{1}{2} (y - ax)^{-1/2}.$$\n\nTherefore,\n$$\frac{\partial^2 z}{\partial y^2} = 2 \sec^2(y + ax) \tan(y + ax) + \frac{3}{4} (y - ax)^{-1/2}.$$\n\n6. **Step 5: Compute the expression $$\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2}.$$**\nSubstitute the values:\n$$= \left(2a^2 \sec^2(y + ax) \tan(y + ax) + \frac{3a^2}{4} (y - ax)^{-1/2} \right) - a^2 \left(2 \sec^2(y + ax) \tan(y + ax) + \frac{3}{4} (y - ax)^{-1/2} \right).$$\n\nDistribute $$a^2$$ in the second term:\n$$= 2a^2 \sec^2(y + ax) \tan(y + ax) + \frac{3a^2}{4} (y - ax)^{-1/2} - 2a^2 \sec^2(y + ax) \tan(y + ax) - \frac{3a^2}{4} (y - ax)^{-1/2}.$$\n\n7. **Step 6: Simplify the expression.**\nThe terms cancel out exactly:\n$$2a^2 \sec^2(y + ax) \tan(y + ax) - 2a^2 \sec^2(y + ax) \tan(y + ax) = 0,$$\n$$\frac{3a^2}{4} (y - ax)^{-1/2} - \frac{3a^2}{4} (y - ax)^{-1/2} = 0.$$\n\nTherefore,\n$$\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2} = 0.$$\n\n**Final answer:** $$\boxed{0}.$$