\(\begin{array}{l}\quad f(x,y)= \dfrac{x^2y}{x^3 + y^3}\\
\quad \text{Ta có:}\\
+)\quad \dfrac{\partial f}{\partial x}= \dfrac{y(2xy^3 - x^4)}{(x^3 + y^3)^2}\\
+)\quad \dfrac{\partial f}{\partial y}= \dfrac{x^2(x^3 - 2y^3)}{(x^3 + y^3)^2}\\
+)\quad \dfrac{\partial^2f}{\partial x^2} = \dfrac{2y^4(-2x^3 - 3x^2 + y^2)}{(x^3 + y^3)^3}\\
+)\quad \dfrac{\partial^2f}{\partial x\partial y}=\dfrac{\partial^2f}{\partial y\partial x}= \dfrac{-x^6 + 5x^4y^3 +8x^3y^3 - 4xy^6}{(x^3 + y^3)^3}\\
+)\quad \dfrac{\partial^2f}{\partial y^2} = - \dfrac{6x^2y^2(x^3 + x^2 - y^3)}{(x^3 + y^3)^3}\\
\text{Tại điểm $(1;1)$ ta được:}\\
+)\quad \dfrac{\partial^2f}{\partial x^2} =-1\\
+)\quad \dfrac{\partial^2f}{\partial x\partial y}=\dfrac{\partial^2f}{\partial y\partial x}=1\\
+)\quad \dfrac{\partial^2f}{\partial y^2} = -\dfrac34\end{array}\)
