The task is to show that the mixed partial derivatives are equal: $f_{xyy} = f_{yyx} = f_{yxy}$, for the function $f(x, y) = \sin(xy)$.

Steps:

1. Find the first partial derivatives:

   • The function is $f(x, y) = \sin(xy)$.
   • First, compute the partial derivatives with respect to $x$ and $y$.

   $f_x = \frac{\partial}{\partial x} \sin(xy) = y \cos(xy)$ $f_y = \frac{\partial}{\partial y} \sin(xy) = x \cos(xy)$

2. Find the second partial derivatives:

   • Compute the second partial derivative with respect to $y$ first.

   $f_{yy} = \frac{\partial}{\partial y} \left( x \cos(xy) \right) = x \left( -x \sin(xy) \right) = -x^2 \sin(xy)$

   • Now compute the mixed partial derivative $f_{xy}$.

   $f_{xy} = \frac{\partial}{\partial y} \left( y \cos(xy) \right) = \cos(xy) - xy \sin(xy)$

3. Find the third mixed partial derivatives:

   • Compute $f_{xyy}$, $f_{yxy}$, and $f_{yyx}$ and show they are equal.

   $f_{xyy} = \frac{\partial}{\partial y} \left( \cos(xy) - xy \sin(xy) \right) = -x \sin(xy) - x \sin(xy) - xy^2 \cos(xy)$ $f_{yxy} = \frac{\partial}{\partial x} \left( -x^2 \sin(xy) \right) = -x^2 \cos(xy) + x^3 \sin(xy)$ [ f_{yyx} = \frac{\partial}{\partial x} \left( -x^3