1. The problem is to evaluate the double integral $$\int_0^1 \int_x^1 y^2 e^{xy} \, dy \, dx.$$\n\n2. We first focus on the inner integral with respect to $y$: $$\int_x^1 y^2 e^{xy} \, dy.$$\n\n3. Since $x$ is treated as a constant in the inner integral, we can use integration by parts or notice the structure. Let's use substitution: define $$u = xy,$$ then $$du = x \, dy,$$ so $$dy = \frac{du}{x}.$$ However, the limits in terms of $u$ are $$u = x \cdot y,$$ so when $y = x$, $$u = x^2,$$ and when $y = 1,$ $$u = x.$$ This substitution is complicated because $x$ varies. So instead, integrate by parts or use the derivative with respect to $y$ approach directly.\n\n4. Consider integrating by parts: set $$v = y^2$$ and $$dw = e^{xy} dy.$$ Then, $$dv = 2y dy$$ and $$w = \frac{1}{x} e^{xy}.$$\n\n5. Applying integration by parts formula: $$\int v \, dw = vw - \int w \, dv,$$ we get\n$$\int_x^1 y^2 e^{xy} dy = \left. y^2 \cdot \frac{e^{xy}}{x} \right|_x^1 - \int_x^1 \frac{e^{xy}}{x} \, 2y dy.$$\n\n6. Simplify the first term: $$\frac{1^2 e^{x \cdot 1}}{x} - \frac{x^2 e^{x \cdot x}}{x} = \frac{e^{x}}{x} - x e^{x^2}.$$\n\n7. The remaining integral is $$\frac{2}{x} \int_x^1 y e^{xy} dy.$$\n\n8. Compute $$\int_x^1 y e^{xy} dy$$ again by parts or directly. Set $$v = y,$$ $$dw = e^{xy} dy,$$ so $$dv = dy,$$ $$w = \frac{e^{xy}}{x}.$$\n\n9. Then, $$\int_x^1 y e^{xy} dy = \left. y \frac{e^{xy}}{x} \right|_x^1 - \int_x^1 \frac{e^{xy}}{x} dy = \frac{e^{x}}{x} - \frac{1}{x} \int_x^1 e^{xy} dy.$$\n\n10. Now compute $$\int_x^1 e^{xy} dy = \left. \frac{e^{xy}}{x} \right|_x^1 = \frac{e^{x} - e^{x^2}}{x}.$$\n\n11. Substitute back to the previous step: $$\int_x^1 y e^{xy} dy = \frac{e^x}{x} - \frac{1}{x} \cdot \frac{e^{x} - e^{x^2}}{x} = \frac{e^{x}}{x} - \frac{e^{x} - e^{x^{2}}}{x^{2}} = \frac{x e^x - (e^x - e^{x^{2}})}{x^{2}} = \frac{x e^{x} - e^{x} + e^{x^{2}}}{x^{2}}.$$\n\n12. Now substitute into step 7: $$\frac{2}{x} \times \int_x^1 y e^{xy} dy = \frac{2}{x} \times \frac{x e^{x} - e^{x} + e^{x^{2}}}{x^{2}} = \frac{2(x e^{x} - e^{x} + e^{x^{2}})}{x^{3}}.$$\n\n13. Putting this into the original inner integral value from step 5, we have\n$$\int_x^1 y^2 e^{xy} dy = \frac{e^{x}}{x} - x e^{x^2} - \frac{2(x e^{x} - e^{x} + e^{x^{2}})}{x^{3}}.$$\n\n14. Simplify the expression to combine terms:\n$$= \frac{e^{x}}{x} - x e^{x^2} - \frac{2x e^{x} - 2 e^{x} + 2 e^{x^{2}}}{x^{3}} = \frac{e^{x}}{x} - x e^{x^{2}} - \frac{2x e^{x}}{x^{3}} + \frac{2 e^{x}}{x^{3}} - \frac{2 e^{x^{2}}}{x^{3}}.$$\n\n15. Rewrite powers of $x$: $$\frac{2x e^{x}}{x^3} = \frac{2 e^{x}}{x^{2}},$$ so\n$$\int_x^1 y^{2} e^{xy} dy = \frac{e^{x}}{x} - x e^{x^{2}} - \frac{2 e^{x}}{x^{2}} + \frac{2 e^{x}}{x^{3}} - \frac{2 e^{x^{2}}}{x^{3}}.$$ \n16. Now, integrate this result with respect to $x$ from 0 to 1: $$\int_0^1 \left(\frac{e^{x}}{x} - x e^{x^{2}} - \frac{2 e^{x}}{x^{2}} + \frac{2 e^{x}}{x^{3}} - \frac{2 e^{x^{2}}}{x^{3}} \right) dx.$$\n\n17. Notice the terms contain singularities at $x=0$, making direct integration difficult. Instead, swap the order of integration. \n18. The original integral is $$\int_0^1 \int_x^1 y^{2} e^{xy} dy dx.$$ Changing order of integration, the limits for $y$ are from 0 to 1, and for each $y$, $x$ goes from 0 to $y$. \n19. So the integral becomes $$\int_0^1 \int_0^y y^{2} e^{xy} dx dy = \int_0^1 y^{2} \left( \int_0^y e^{xy} dx \right) dy.$$\n\n20. Compute the inner integral with respect to $x$:\n$$\int_0^y e^{xy} dx = \left. \frac{1}{y} e^{xy} \right|_0^y = \frac{e^{y^{2}} - 1}{y}.$$\n\n21. Substitute back to have\n$$\int_0^1 y^{2} \cdot \frac{e^{y^{2}} - 1}{y} dy = \int_0^1 y (e^{y^{2}} - 1) dy = \int_0^1 y e^{y^{2}} dy - \int_0^1 y dy.$$\n\n22. Make the substitution $$u = y^{2},$$ so $$du = 2 y dy,$$ implying $$y dy = \frac{du}{2}.$$\n\n23. The first integral becomes \n$$\int_0^1 y e^{y^{2}} dy = \int_0^1 e^{u} \frac{du}{2} = \frac{1}{2} \int_0^1 e^{u} du = \frac{1}{2} (e^{1} - e^{0}) = \frac{e - 1}{2}.$$\n\n24. The second integral is straightforward: $$\int_0^1 y dy = \frac{1^2}{2} = \frac{1}{2}.$$\n\n25. Combine the results: \n$$\frac{e - 1}{2} - \frac{1}{2} = \frac{e - 2}{2}.$$\n\n26. Therefore, the value of the double integral is $$\boxed{\frac{e - 2}{2}}.$$