1. The expression provided is $((x^2 - y^2) + 1 (x y^2))$. Let's clarify and rewrite it for proper interpretation: $$x^2 - y^2 + xy^2$$ 2. We can analyze the function $f(x,y) = x^2 - y^2 + xy^2$ which combines quadratic terms in both $x$ and $y$. 3. The graph described appears to be related to the level curves or the implicit plot of $f(x,y)$, showing bracket-like curves oriented vertically. 4. To understand the geometry, consider setting $f(x,y) = c$ for constant $c$, which forms curves: $$x^2 - y^2 + xy^2 = c$$ 5. These curves will have asymmetry due to the mixed term $xy^2$, resulting in vertical bracket-like shapes concentrated on the left side. 6. To analyze intercepts, set $y=0$: $$f(x,0) = x^2 = c$$ so along the $x$-axis, the function is just $x^2$. 7. Similarly, for $x=0$: $$f(0,y) = -y^2 = c$$ which are downward opening parabolas along the $y$-axis. 8. Extrema and critical points can be found by partial derivatives: $$\frac{\partial f}{\partial x} = 2x + y^2$$ $$\frac{\partial f}{\partial y} = -2y + 2xy$$ 9. Setting both partial derivatives to zero: $$2x + y^2 = 0$$ $$-2y + 2xy = 0 \implies 2y(x - 1) = 0$$ 10. From $2y(x-1) = 0$, either $y=0$ or $x=1$. 11. If $y=0$, from $2x + y^2=0$, $2x=0$ so $x=0$. 12. If $x=1$, from $2x + y^2=0$: $$2(1) + y^2=0 \implies 2 + y^2=0 \implies y^2 = -2$$ which is not possible for real $y$. 13. Therefore, critical point at $(0,0)$. 14. Evaluate $f(0,0) = 0$. 15. To determine nature of critical point, consider second derivatives: $$f_{xx} = 2, \quad f_{yy} = -2 + 2x, \quad f_{xy} = 2y$$ 16. At $(0,0)$: $$f_{xx} = 2, \quad f_{yy} = -2, \quad f_{xy} = 0$$ 17. The Hessian determinant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = 2 \times (-2) - 0 = -4 < 0$$ 18. Negative Hessian determinant indicates a saddle point at $(0,0)$. **Final summary:** The function $$f(x,y) = x^2 - y^2 + xy^2$$ has a saddle point at the origin. The graph consists of three vertical bracket-like curves primarily located on the left side due to the mixed term $xy^2$. The curves are implicit solutions to $$x^2 - y^2 + xy^2 = c.$$