1. **State the problem:** Solve the partial differential equation (PDE) $$x^2 \frac{\partial^2 z}{\partial x^2} - y^2 \frac{\partial^2 z}{\partial y^2} = xy.$$\n\n2. **Identify the PDE type and approach:** This is a second-order PDE with mixed variables. The equation can be written as $$x^2 z_{xx} - y^2 z_{yy} = xy,$$ where $z_{xx} = \frac{\partial^2 z}{\partial x^2}$ and $z_{yy} = \frac{\partial^2 z}{\partial y^2}$.\n\n3. **Try a particular solution:** Since the right side is $xy$, try a polynomial form $z = Axy$ where $A$ is a constant to be determined.\n\n4. **Compute derivatives:**\n- $z_x = A y$\n- $z_{xx} = 0$\n- $z_y = A x$\n- $z_{yy} = 0$\n\nSubstitute into the PDE: $$x^2 \cdot 0 - y^2 \cdot 0 = xy \implies 0 = xy,$$ which is false, so $z = Axy$ is not a particular solution.\n\n5. **Try a more general form:** Assume $z = f(x)g(y)$. Then\n$$z_{xx} = f''(x)g(y), \quad z_{yy} = f(x)g''(y).$$\nSubstitute into PDE:\n$$x^2 f''(x) g(y) - y^2 f(x) g''(y) = xy.$$\nDivide both sides by $f(x)g(y)$ (assuming nonzero):\n$$x^2 \frac{f''(x)}{f(x)} - y^2 \frac{g''(y)}{g(y)} = \frac{xy}{f(x)g(y)}.$$\nThe right side depends on both $x$ and $y$ separately, so separation of variables is complicated.\n\n6. **Try a solution of the form:** $$z = X(x)Y(y) + Z_p(x,y),$$ where $Z_p$ is a particular solution.\n\n7. **Find a particular solution:** Try $z_p = Bxy$, compute derivatives as before, which yield zero on the left side, so no. Try $z_p = C x^m y^n$.\n\nCompute derivatives:\n$$z_{xx} = C m (m-1) x^{m-2} y^n, \quad z_{yy} = C n (n-1) x^m y^{n-2}.$$\nSubstitute into PDE:\n$$x^2 C m (m-1) x^{m-2} y^n - y^2 C n (n-1) x^m y^{n-2} = xy,$$\nwhich simplifies to\n$$C m (m-1) x^m y^n - C n (n-1) x^m y^n = xy,$$\nor\n$$C [m(m-1) - n(n-1)] x^m y^n = xy.$$\n\nFor this to equal $xy$, we need $m=1$, $n=1$, and\n$$C [1(0) - 1(0)] = C \cdot 0 = 1,$$ which is impossible.\n\n8. **Try a solution of the form:** $$z = A x y \ln(x) + B x y \ln(y).$$\nCompute derivatives and substitute to check if this works.\n\n9. **Compute derivatives for $z = x y \ln x$:**\n$$z_x = y (\ln x + 1), \quad z_{xx} = y \frac{1}{x},$$\nso\n$$x^2 z_{xx} = x^2 y \frac{1}{x} = x y x = x^2 y.$$\nSimilarly for $z = x y \ln y$,\n$$z_y = x (\ln y + 1), \quad z_{yy} = x \frac{1}{y},$$\nso\n$$- y^2 z_{yy} = - y^2 x \frac{1}{y} = - x y y = - x y^2.$$\n\n10. **Combine:**\n$$x^2 z_{xx} - y^2 z_{yy} = x^2 y - x y^2 = x y (x - y).$$\nWe want $xy$, so try a linear combination:\n$$z = A x y \ln x + B x y \ln y,$$\nthen\n$$x^2 z_{xx} - y^2 z_{yy} = A x y x - B x y y = x y (A x - B y).$$\nSet equal to $xy$, so\n$$x y (A x - B y) = x y,$$\nwhich implies\n$$A x - B y = 1$$\nfor all $x,y$, impossible unless $A = B = 0$.\n\n11. **Conclusion:** The homogeneous equation is\n$$x^2 z_{xx} - y^2 z_{yy} = 0,$$\nwhich is separable. The general solution is the sum of the homogeneous solution plus a particular solution.\n\n12. **Solve homogeneous PDE:** Assume $z = X(x) Y(y)$, then\n$$x^2 X'' Y - y^2 X Y'' = 0 \implies \frac{x^2 X''}{X} = \frac{y^2 Y''}{Y} = \lambda,$$ a separation constant.\n\n13. **Solve ODEs:**\n$$x^2 X'' - \lambda X = 0, \quad y^2 Y'' - \lambda Y = 0.$$\nThese are Euler-Cauchy equations.\n\n14. **General solution for $X$:** Try $X = x^m$, then\n$$x^2 m (m-1) x^{m-2} - \lambda x^m = 0 \implies m(m-1) - \lambda = 0,$$\nso\n$$m = \frac{1 \pm \sqrt{1 + 4 \lambda}}{2}.$$\nSimilarly for $Y = y^n$,\n$$n(n-1) - \lambda = 0,$$\nso\n$$n = \frac{1 \pm \sqrt{1 + 4 \lambda}}{2}.$$\n\n15. **Homogeneous solution:**\n$$z_h = \sum \left( A_m x^{m} y^{n} + B_m x^{m} y^{n'} \right),$$\nwhere $m,n$ satisfy the above relations.\n\n16. **Particular solution:** Try $z_p = K x y$, compute derivatives:\n$$z_{xx} = 0, \quad z_{yy} = 0,$$ so no contribution. Try $z_p = K x^a y^b$ with $a,b$ to be found.\n\n17. **Try $a=1$, $b=1$ fails, try $a=2$, $b=1$: $$z_p = K x^2 y.$$\nThen\n$$z_{xx} = K 2 (2-1) x^{0} y = 2 K y,$$\n$$z_{yy} = 0,$$\nso\n$$x^2 z_{xx} - y^2 z_{yy} = x^2 (2 K y) - 0 = 2 K x^2 y.$$\nSet equal to $xy$,\n$$2 K x^2 y = x y \implies 2 K x = 1,$$ which is not true for all $x$.\n\n18. **Try $a=1$, $b=2$: $$z_p = K x y^2.$$\nThen\n$$z_{xx} = 0,$$\n$$z_{yy} = K x 2 (2-1) y^{0} = 2 K x,$$\nso\n$$x^2 z_{xx} - y^2 z_{yy} = 0 - y^2 (2 K x) = - 2 K x y^2.$$\nSet equal to $xy$,\n$$- 2 K x y^2 = x y \implies - 2 K y = 1,$$ not true for all $y$.\n\n19. **Try $z_p = K x y$ times a function of $x/y$ or $y/x$ or logarithms, but this is complex.**\n\n20. **Summary:** The general solution is the sum of the homogeneous solution (Euler-Cauchy type) plus a particular solution which may require advanced methods (variation of parameters or integral transforms).\n\n**Final answer:**\n$$z = \sum \left( A_m x^{m} y^{n} + B_m x^{m} y^{n'} \right) + z_p,$$\nwhere $m,n$ satisfy $m(m-1) = n(n-1) = \lambda$, and $z_p$ is a particular solution to $$x^2 z_{xx} - y^2 z_{yy} = xy.$$