Find \(f_x\), \(f_y\), \(f_z\), \(f_{yz}\), and \(f_{zy}\).
\(f(x,y,z)=x^3y^2 +x^3z+y^2z\)
\[ f_x = \frac{\partial}{\partial x} (x^3 y^2 + x^3 z + y^2 z)\\ f_x = 3x^2 y^2 + 3x^2 z\\ f_y = \frac{\partial}{\partial y}(x^3 y^2 + x^3 z + y^2 z)\\ f_y = 2x^3 y + 2yz\\ f_z = \frac{\partial}{\partial z}(x^3 y^2 + x^3 z + y^2 z)\\ f_z = x^3 + y^2\\ f_{yz} = \frac{\partial^2}{\partial y \partial z} (x^3 y^2 + x^3 z + y^2 z)\\ \text{First derive with respect to y, and then with respect to z:}\\ = \frac{\partial}{\partial z} (2yx^3 + 2yz) = \\ = \frac{\partial}{\partial z} (2y(x^3 + z)) = 2y (0 + 1 ) = 2y\\ f_{yz}= 2y\\ \text{First derive with respect to z, and then with respect to y:}\\ f_{zy} = \frac{\partial}{\partial z \partial y} (x^3 y^2 + x^3 z + y^2 z) = \frac{\partial}{\partial y} (x^3 + y^2) = 2y\\ f_{zy} = 2y\\ \]
The solutions are as follows:
\[ f_x = 3x^2 y^2 + 3x^2 z\\ f_y = 2x^3 y + 2yz\\ f_z = x^3 + y^2\\ f_{yz}= 2y\\ f_{zy} = 2y\\ \]